Algebra II (MA249) Lecture Notes

Derek Holt

Contents

I Group Theory 3

1 Definition, Examples and Elementary Properties 3

1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Examples – Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Examples – Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Examples – Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5 Elementary Properties – the Cancellation Laws . . . . . . . . . . . . . . . . . 5

1.6 Elementary Properties – Orders of Elements . . . . . . . . . . . . . . . . . . . 5

1.7 Examples – Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.8 Isomorphisms and Isomorphic groups . . . . . . . . . . . . . . . . . . . . . . . 6

1.9 Direct Products and Groups of Order 4 . . . . . . . . . . . . . . . . . . . . . 7

1.10 Examples – the Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.11 Generators and Defining Relations . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Subgroups of Groups 10

2.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 Cosets and Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Normal Subgroups and Quotient Groups 14

3.1 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2 Classification of Groups of Orders 6 and 8 . . . . . . . . . . . . . . . . . . . . 14

3.3 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.4 Examples of Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Homomorphisms and the Isomorphism Theorems 17

4.1 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.2 Kernels and Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.3 The Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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5 Groups Acting On Sets 21

5.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.2 Orbits and Stabilisers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.3 Conjugacy Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.4 The Simplicity of A5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.5 Sylow’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

II Ring Theory 28

6 Definition, Examples and Elementary Properties 28

6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

6.1.1 Examples – Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.1.2 Examples – Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.1.3 Examples – Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.2 Subrings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6.3 Isomorphisms and direct products . . . . . . . . . . . . . . . . . . . . . . . . 30

7 Homomorphisms, Ideals, Quotient Rings and the Isomorphism Theorems 31

7.1 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

7.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7.3 Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7.4 Maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

8 Euclidean Domains, Principal Ideal Domains, and Unique FactorisationDomains 34

8.1 Euclidean and Principal Ideal Domains . . . . . . . . . . . . . . . . . . . . . . 34

8.2 Prime and Irreducible Elements . . . . . . . . . . . . . . . . . . . . . . . . . . 36

8.3 Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

8.4 Unique Factorisation Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

8.5 Primes in F [x] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8.6 Gaussian primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

9 Polynomial Rings over UFDs 43

9.1 Fields of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

9.2 Polynomial rings over UFDs are UFDs . . . . . . . . . . . . . . . . . . . . . . 45

9.3 The cyclotomic polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

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Part I

Group Theory

1 Definition, Examples and Elementary Properties

1.1 Definitions

Definition A group is a set G together with a binary operation ◦ : G×G→ G that satisfiesthe following properties:(i) (Closure) For all g, h ∈ G, g ◦ h ∈ G;(ii) (Associativity) For all g, h, k ∈ G, (g ◦ h) ◦ k = g ◦ (h ◦ k);(iii) There exists an element e ∈ G such that:

(a) (Identity) for all g ∈ G, e ◦ g = g; and(b) (Inverse) for all g ∈ G there exists h ∈ G such that h ◦ g = e.

(Property (i) does not really need stating, because it is implied by the fact that ◦ : G×G→ Gis a binary operation on G. But it is traditionally the first of the four group axioms, so wehave included it here!)

The number of elements in G is called the order of G and is denoted by |G|. This may befinite or infinite.

An element e ∈ G satisfying (iii) of the definition is called an identity element of G, and forg ∈ G, an element h that satisfies (iii)(b) of the definition (h ◦ g = e) is called an inverseelement of g.

We shall immediately prove two technical lemmas, which are often included as part of thedefinition of a group. These two proofs need not be memorised!

Lemma 1.1 Let G be a group, let e ∈ G be an identity element, and for g ∈ G, let h ∈ G bean inverse element of g. Then g ◦ e = g and g ◦ h = e.

Proof: We have h ◦ (g ◦ e) = (h ◦ g) ◦ e = e ◦ e = e = h ◦ g. Now let h′ be an inverseof h. Then multiplying the left and right sides of this equation on the left by h′ and usingassociativity gives (h′ ◦ h) ◦ (g ◦ e) = (h′ ◦ h) ◦ g. But (h′ ◦ h) ◦ (g ◦ e) = e ◦ (g ◦ e) = g ◦ e, and(h′ ◦ h) ◦ g = e ◦ g = g, so we get g ◦ e = g.

We have h ◦ (g ◦ h) = (h ◦ g) ◦ h = e ◦ h = h, and multiplying on the left by h′ gives(h′ ◦h)◦ (g ◦h) = h′ ◦h. But (h′ ◦h)◦ (g ◦h) = e◦ (g ◦h) = g ◦h and (h′ ◦h) = e, so g ◦h = e.2

Lemma 1.2 Let G be a group. Then G has a unique identity element, and any g ∈ G has aunique inverse.

Proof: Let e and f be two identity elements of G. Then, e ◦ f = f , but by Lemma 1.1, wealso have e ◦ f = e, so e = f and the identity element is unique.

Let h and h′ be two inverses for g. Then h ◦ g = h′ ◦ g = e, but by Lemma 1.1 we also haveg ◦ h = e, so

h = e ◦ h = (h′ ◦ g) ◦ h = h′ ◦ (g ◦ h) = h′ ◦ e = h′

and the inverse of g is unique. 2

Definition A group is called abelian or commutative if it satisfies the additional property:

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(Commutativity) For all g, h ∈ G, g ◦ h = h ◦ g.We shall now proceed to change notation!

The groups in this course will either be:• Multiplicative groups, where we omit the ◦ sign (g ◦ h becomes just gh), we denote the

identity element by 1 rather than by e, and we denote the inverse of g ∈ G by g−1; or• Additive groups, where we replace ◦ by +, we denote the identity element by 0, and we

denote the inverse of g by −g.

If there is more than one group around, and we need to distinguish between the identityelements of G and H say, then we will denote them by 1G and 1H (or 0G and 0H).

Additive groups will always be commutative, but multiplicative groups may or may not becommutative. The default will be to use the multiplicative notation.

The proof of the next lemma is easy and is left as an exercise. From now on, this result willbe used freely and without explicit reference.

Lemma 1.3 Let g, h be elements of the multiplicative group G. Then (gh)−1 = h−1g−1.

1.2 Examples – Numbers

Z, Q, R, and C or indeed the elements of any field form a group under addition. We sometimesdenote these by (Z,+), (Q,+), etc.

Now let K be any field, such as Q, R or C, and let K∗ = K \ {0}. Then K∗ is a group undermultiplication. But note that Z∗ = Z\{0} is not a group under multiplication, because mostelements do not have multiplicative inverses.

All of these groups are abelian.

1.3 Examples – Matrix Groups

Let K be a field. Then, for any fixed n > 0, the set of n× n invertible matrices with entriesin K forms a group under multiplication. This group is denoted by GL(n,K) or GLn(K).

The n × n matrices over K with determinant 1 also form a group denoted by SL(n,K) orSLn(K). This is a subgroup of GL(n,K).

Another subgroup of GL(n,K) is the group O(n,K), which consists of the orthogonal n× nmatrices over K; that is, those matrices A such that AT = A−1. (Proof left as exercise.)

These groups are not usually abelian. In fact, except for a couple of examples when K is asmall finite field, they are abelian only when n = 1.

1.4 Examples – Permutation Groups

Let X be any set, and let Sym(X) denote the set of permutations of X; that is, the bijectionsfrom X to itself. Then Sym(X) is a group under composition of maps. It is known as thesymmetric group on X.

The proof that Sym(X) is a group uses results from Foundations. Note that the compositionof two bijections is a bijection, and that composition of any maps obeys the associative law.The identity element of the group is just the identity map X → X, and the inverse elementof a map is just its inverse map.

Let us recall the cyclic notation for permutations. If a1, . . . , ar are distinct elements of X,then the cycle (a1, a2, . . . , ar) denotes the permutation of φ ∈ X with(i) φ(ai) = ai+1 for 1 ≤ i < r.

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(ii) φ(ar) = a1, and(iii) φ(b) = b for b ∈ X \ {a1, a2, . . . , ar}.

When X is finite, any permutation of X can be written as a product (= composite) of disjointcycles. Note that a cycle (a1) of length 1 means that φ(a1) = a1, and so this cycle can (andnormally is) omitted.

For example, if X = {1, 2, 3, 4, 5, 6, 7, 8} and φ maps 1, 2, 3, 4, 5, 6, 7, 8 to 5, 8, 6, 4, 3, 1, 2, 7,respectively, then φ = (1, 5, 3, 6)(2, 8, 7), where the cycle (4) of length 1 has been omitted.We will denote the identity permutation in cyclic notation by ().

Remember that a composite φ2φ1 of maps means φ1 followed by φ2, so, for example, ifX = {1, 2, 3}, φ1 = (1, 2, 3) and φ2 = (1, 2), then φ1φ2 = (1, 3), whereas φ2φ1 = (2, 3). Thisexample shows that Sym(X) is not in general a commutative group. (In fact it is commutativeonly when |X| ≤ 2.)

The inverse of a permutation can be calculated easily by just reversing all of the cycles.For example, the inverse of (1, 5, 3, 6)(2, 8, 7) is (6, 3, 5, 1)(7, 8, 2), which is the same as(1, 6, 3, 5)(2, 7, 8). (The cyclic representation is not unique: (a1, a2, . . . , ar) = (a2, a3, . . . , ar, a1),etc.)

1.5 Elementary Properties – the Cancellation Laws

Proposition 1.4 Let G be any group, and let g, h, k ∈ G. Then(i) gh = gk ⇒ h = k; and(ii) hg = kg ⇒ h = k.

Proof: For (i), we have gh = gk ⇒ g−1gh = g−1gk ⇒ h = k, and (ii) is proved similarly bymultiplying by g−1 on the right. 2

1.6 Elementary Properties – Orders of Elements

First some more notation. In a multiplicative group G, we define g2 = gg, g3 = ggg,g4 = gggg, etc. Formally, for n ∈ N, we define gn inductively, by g1 = g and gn+1 = ggn forn ≥ 1. We also define g0 to be the identity element 1, and g−n to be the inverse of gn. Thengm+n = gmgn for all m,n ∈ Z.

In an additive group, gn becomes ng, where 0g = 0, and (−n)g = −(ng).

Definition Let g ∈ G. Then the order of g, denoted by o(g) or by |g|, is the least n > 0 suchthat gn = 1, if such an n exists. If there is no such n, then g has infinite order, and we write|g| = ∞.

Note that if g has infinite order, then the elements gk are distinct for distinct values of k,because if gj = gk with j < k, then gk−j = 1 and g has finite order.

Similarly, if g has finite order n, then the n elements g0 = 1, g1 = g, . . . , gn−1 = g−1 are alldistinct, and for any k ∈ Z, gk is equal to exactly one of these n elements.

Lemma 1.5 |g| = 1 ⇔ g = 1.

Lemma 1.6 If |g| = n then, for k ∈ Z, gk = 1 ⇔ n|k.

(Recall notation: for integers j, k, j|k means j divides k.)

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1.7 Examples – Cyclic Groups

Definition A group G is called cyclic, if it consists of the integral powers of a single element.In other words, G is cyclic if there exists an element g in G with the property that, for allh ∈ G, there exists k ∈ Z with gk = h. The element g is called a generator of G.

The most familiar examples of cyclic groups are additive groups rather than multiplicative.The group (Z,+) of integers under addition is cyclic, because every integer k is equal to k1,and so 1 is a generator.

If we fix some integer n > 0 and let Zn = {0, 1, 2, . . . , n−1} with the operation of additionmodulo n, then we get a cyclic group of order n, where 1 is once again a generator. (Is 1 theonly generator? If not, which elements in Zn are generators?)

Lemma 1.7 In an infinite cyclic group, any generator g has infinite order. In a finite cyclicgroup of order n, any generator g has order n.

1.8 Isomorphisms and Isomorphic groups

Later on (Section 4), we shall be considering the more general case of homomorphisms betweengroups, but for now we just introduce the important special case of isomorphisms.

Definition An isomorphism φ : G → H between two groups G and H is a bijection fromG to H such that φ(g1g2) = φ(g1)φ(g2) for all g1, g2 ∈ G. Two groups G and H are calledisomorphic if there is an isomorphism between them. In this case we write G ∼= H.

The reason this is so important is that isomorphic groups are considered to be essentially thesame group – H can be obtained from G simply be relabelling the elements of G.

Exercise Show that the relationship between groups of being isomorphic satisfies the condi-tions of an equivalence relation; that is, G ∼= G, G ∼= H ⇒ H ∼= G, and G ∼= H,H ∼= K ⇒G ∼= K.

The following proposition says that there is essentially only one cyclic group of each finiteorder, and one of infinite order. For this reason it is customary to talk about the cyclic groupof order n, and the infinite cyclic group.

Proposition 1.8 Any two infinite cyclic groups are isomorphic. For a positive integer n,any two cyclic groups of order n are isomorphic.

Proof: If G and H are infinite cyclic groups with generators g and h, then G = {gk | k ∈ Z}and H = {hk | k ∈ Z}. We saw in Subsection 1.7 that the elements gk of G are distinct fordistinct k ∈ Z, and so the map φ : G→ H defined by φ(gk) = hk for all k ∈ Z is a bijection,and it is easily checked to be an isomorphism.

If G and H are finite cyclic groups of order n, then G = {gk | k ∈ Zn} and H = {hk | k ∈ Zn}and the map φ : G→ H defined by φ(gk) = hk for all k ∈ Zn is an isomorphism. 2

We denote a cyclic group of order m by Cm. Since any two such groups are isomorphic, thisnotation is effectively unambiguous.

Here is another easy example of a group isomorphism.

Proposition 1.9 Let X and Y be two sets with |X| = |Y |. Then the groups Sym(X) andSym(Y ) of all permutations of X and Y are isomorphic.

The notation Sym(n) or Sn is standard for the symmetric group on a set X with |X| = n.By default, we take X = {1, 2, 3, . . . , n}.

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The following result is often useful; its proof is left as an exercise.

Proposition 1.10 If φ : G→ H is an isomorphism, then |g| = |φ(g)| for all g ∈ G.

1.9 Direct Products and Groups of Order 4

Definition Let G and H be two (multiplicative) groups. We define the direct product G×Hof G and H to be the set {(g, h) | g ∈ G, h ∈ H} of ordered pairs of elements from G andH, with the obvious component-wise multiplication of elements (g1, h1)(g2, h2) = (g1g2, h1h2)for g1, g2 ∈ G and h1, h2 ∈ H.

It is straightforward to check that G × H is a group under this operation. Note that theidentity element is (1G, 1H), and the inverse of (g, h) is just (g−1, h−1).

If the groups are additive, then it is usually called the direct sum rather than the directproduct, and written G⊕H.

A large part of group theory consists of classifying groups with various properties. Thismeans finding representatives of the isomorphism classes of groups with these properties. Asan example of this, let us consider groups of order 4. Let G = {1, a, b, c} be such a group.

Let us consider the orders of the elements a, b and c. These orders cannot be more than 4,because |G| = 4, and they cannot be 1 by Lemma 1.5. This leaves 2, 3 and 4 as possibilities.If an element, a, say, has order 4, then G = {1, a, a2, a3} is a cyclic group.

Order 3 is impossible. We shall prove later in Proposition 2.11 that the order of an elementmust divide the order of the group, but we can see this directly by using the cancellation laws(Proposition 1.4). If |a| = 3, then we can assume that a2 = b, and a3 = 1, so G = {1, a, a2, c}.Now what can ac be? ac = a ⇒ ac = a1 ⇒ c = 1: wrong! ac = c ⇒ ac = 1c ⇒ a = 1:wrong! ac = a2 ⇒ a = c: wrong! ac = 1 ⇒ ac = a3 ⇒ c = a2: wrong!

Thus either G is cyclic, or a, b and c all have order 2. In that case, we can use the cancellationlaws to prove that the product of any two of a, b, cmust be the third; for example, ab = ba = c,ca = b, etc. In fact we can write its complete multiplication table:

1 a b c

1 1 a b ca a 1 c bb b c 1 ac c b a 1

Since we have completely determined the multiplication table, we can say immediately thatany two groups of order four in which the non-identity elements all have order 2 are iso-morphic. For example, the direct product C2 × C2 of two cyclic groups of order 2 has thisproperty. This group is known as the Klein Four Group. Summing up, we have:

Proposition 1.11 A group of order 4 is isomorphic either to a cyclic group or to a KleinFour Group.

Exercise Use Proposition 1.10 to show that a cyclic group of order 4 is not isomorphic to aKlein Four Group.

1.10 Examples – the Dihedral Groups

Let n ∈ N with n ≥ 2 and let P be a regular n-sided polygon in the plane. The dihedralgroup of order 2n consists of the isometries of P . These consist of

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(i) n rotations through the angles 2πk/n (0 ≤ k < n) about the centre of P ; and(ii) n reflections about lines that pass through the centre of P , and either pass through a

vertex of P or bisect an edge of P (or both).

Unfortunately, some books denote this group by Dn and others by D2n, which can be con-fusing! We shall use D2n.

To analyse these groups, it will be convenient to number the vertices in order 1, 2, . . . , n, andto regard the group elements as permutations of these vertices. Then the n rotations are thepowers ak for 0 ≤ k < n, where a = (1, 2, 3, . . . , n) is a rotation through the angle 2π/n,

Let b be the reflection through the bisector of P that passes through the vertex 1. Then binterchanges the vertices 2 and n that are adjacent to 1, and similarly it interchanges 3 andn−1, 4 and n−2, etc., so we can have b = (2, n)(3, n−1)(4, n−2) . . .. For example, when n = 5,b = (2, 5)(3, 4) and when n = 6, b = (2, 6)(3, 5). Notice that there is a difference between theodd and even cases. When n is odd, b fixes no vertex other than 1, but when n is even, bfixes one other vertex, namely (n+2)/2.

Now we can see, either geometrically or by multiplying permutations, that the n reflectionsof P are the elements akb for 0 ≤ k < n. (Remember that ab means first do b and then doa.) Thus we have

G = {ak | 0 ≤ k < n} ∪ {akb | 0 ≤ k < n} (†).

Again, the odd and even cases are slightly different. When n = 5, b, ab, a2b, a3b, a4b areequal to (2, 5)(3, 4), (1, 2)(3, 5), (1, 3)(4, 5), (1, 4)(2, 3), (1, 5)(2, 4) respectively, which are thereflections through the bisectors of P through vertices 1, 4, 2, 5, 3. When n = 6, we haveb = (2, 6)(3, 5), a2b = (1, 3)(4, 6), a4b = (1, 5)(2, 4), which are reflections through bisectorsof P passing through two vertices, whereas ab = (1, 2)(3, 6)(4, 5), a3b = (1, 4)(2, 3)(5, 6),a5b = (1, 6)(2, 5)(3, 4), which are reflections through lines that bisect two edges of P .

In all cases, we have ba = an−1b (= a−1b); this is the reflection that interchanges vertices iand n+1−i for 1 ≤ i ≤ n. Hence we also have bak = an−kb (= a−kb) for 0 ≤ k < n. Theseequations, together with an = 1 and b2 = 1 enable us to calculate the full multiplication tableof G with the elements written as they are in the expression (†) above, because they enableus to perform any of the four basic types of products:(i) (ak)(al) = ak+l (k + l < n) or ak+l−n (k + l ≥ n);(ii) (ak)(alb) = ak+lb (k + l < n) or ak+l−nb (k + l ≥ n);(iii) (akb)(al) = akan−lb = ak+n−lb (k < l) or ak−lb (k ≥ l);(iv) (akb)(alb) = akan−lbb = ak+n−l (k < l) or ak−l (k ≥ l).

Let us write out the full multiplication table in the case n = 6.

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1 a a2 a3 a4 a5 b ab a2b a3b a4b a5b

1 1 a a2 a3 a4 a5 b ab a2b a3b a4b a5ba a a2 a3 a4 a5 1 ab a2b a3b a4b a5b ba2 a2 a3 a4 a5 1 a a2b a3b a4b a5b b aba3 a3 a4 a5 1 a a2 a3b a4b a5b b ab a2ba4 a4 a5 1 a a2 a3 a4b a5b b ab a2b a3ba5 a5 1 a a2 a3 a4 a5b b ab a2b a3b a4bb b a5b a4b a3b a2b ab 1 a5 a4 a3 a2 aab ab b a5b a4b a3b a2b a 1 a5 a4 a3 a2

a2b a2b ab b a5b a4b a3b a2 a 1 a5 a4 a3

a3b a3b a2b ab b a5b a4b a3 a2 a 1 a5 a4

a4b a4b a3b a2b ab b a5b a4 a3 a2 a 1 a5

a5b a5b a4b a3b a2b ab b a5 a4 a3 a2 a 1

1.11 Generators and Defining Relations

Definition The elements {g1, g2, . . . , gr} of a group G are said to generate G (or to form aset of generators for G) if every element of G can be obtained by repeated multiplication ofthe gi and their inverses.

This means that every element of G can be written as an expression like g22g

−12 g1g4g

−13 g1g

−12

in the gi and g−1i , which is allowed to be as long as you like. Such an expression is also called

a word in the generators gi (and their inverses).

Examples 1. A group is cyclic if and only if it can be generated by a single element.

2. The dihedral groups are generated by a = (1, 2, . . . , n) and b = (2, n)(3, n−1)....

3. We shall see in Lemma 2.3 below that, when X is finite, any permutation in the symmetricgroups Sym(X) can be written as a product of transpositions. So the set of all transpositionsgenerates Sym(X). (In fact there are much smaller generating sets for Sym(X).)

We shall be needing some techniques to prove isomorphisms between groups satisfying variousproperties. The following results provide methods of doing this for certain families of groups.

Proposition 1.12 Let G be a group of order 2n generated by two elements a and b thatsatisfy the equations an = 1, b2 = 1 and ba = a−1b. Then G ∼= D2n.

Proof: Since G is generated by a and b, any element of G can be written as a product of thegenerators a, b, a−1, b−1. Since an = 1 and b2 = 1, we can always replace a−1 by an−1 andb−1 by b, so we can assume that only a and b appear in this product. Furthermore, we canuse the equation ba = a−1b = an−1b to move all occurrences of a in the product to the leftof the expression, and we end up with a word in the form akbl. Using an = b2 = 1 again, wecan assume that 0 ≤ k < n and 0 ≤ l < 2. This leaves us with precisely 2n different wordsakbl, and since we are told that |G| = 2n, these words must all represent distinct elements ofG. We have now shown that G = {ak | 0 ≤ k < n} ∪ {akb | 0 ≤ k < n}, exactly as in D2n.

Using ba = a−1b twice, we get ba2 = (ba)a = a−1ba = a−1a−1b = a−2b, and similarlybak = a−kb for all k ≥ 0, and since a−k = an−k, we have bak = an−kb for 0 ≤ k < n. We sawin Subsection 1.10, that these equations, together with an = 1 and b2 = 1 allow us to deducethe whole of the multiplication table of D2n. Hence any group satisfying the hypotheses ofthis proposition has corresponding elements and the same multiplication as D2n, and so it isisomorphic to D2n. 2

The equations {an = 1, b2 = 1, ba = a−1b} are called defining relations for D2n, which meansroughly that D2n is the largest group generated by two elements a and b that satisfy these

9

equations. We shall not go into the general theory of defining relations in this course, but weshall give two more examples.

Consider the direct product Cm × Cn of cyclic groups of orders m and n, which has ordermn. Let c and d be generators of Cm and Cn, and let a = (c, 1) and b = (1, d). Then we caneasily check that am = 1, bn = 1 and ab = ba.

Proposition 1.13 Let G be a group of order mn generated by two elements a and b thatsatisfy the equations am = 1, bn = 1 and ba = ab. Then G ∼= Cm × Cn.

Proof: By a similar argument to that used in the proof of Proposition 1.12, we can showthat any element of G can be written as akbl with 0 ≤ k < m, 0 ≤ l < n, and hence deducethat G = {akbl | 0 ≤ k < m, 0 ≤ l < n}.From ba = ab, we deduce blak = akbl for k, l ≥ 0, and the defining relations now enable useasily to deduce the complete multiplication table of G. The result follows. 2

As a final example, we consider a specific group of order 8, which is known as the quaterniongroup, and denoted by Q8. There are several way to define this. One way is as the subgroupof GL(2,C) consisting of the 8 matrices

1 =(

1 00 1

), a =

(0 1

−1 0

), a2 =

(−1 0

0 −1

), a3 =

(0 −11 0

),

b =(i 00 −i

), ab =

(0 −i−i 0

), a2b =

(−i 00 i

), a3b =

(0 ii 0

).

We have denoted the identity element by 1, as usual, but do not confuse this with the element1 ∈ C !

We can check directly that a4 = 1, b2 = a2, and ba = a3b = a−1b. By a similar argument tothat which we used for the dihedral groups, we can deduce the full multiplication table of Q8

from these equations, and thus prove:

Proposition 1.14 Let G be a group of order 8 generated by two elements a and b that satisfythe equations a4 = 1, b2 = a2 and ba = a−1b. Then G ∼= Q8.

The multiplication table of Q8 is as follows. Notice that it has only a single element (a2) oforder 2, and 6 elements of order 4.

1 a a2 a3 b ab a2b a3b

1 1 a a2 a3 b ab a2b a3ba a a2 a3 1 ab a2b a3b ba2 a2 a3 1 a a2b a3b b aba3 a3 1 a a2 a3b b ab a2bb b a3b a2b ab a2 a 1 a3

ab ab b a3b a2b a3 a2 a 1a2b a2b ab b a3b 1 a3 a2 aa3b a3b a2b ab b a 1 a3 a2

2 Subgroups of Groups

2.1 Definition and Examples

Definition A subset H of a group G is called a subgroup of G if it forms a group under thesame operation as that of G.

10

Lemma 2.1 If H is a subgroup of G, then the identity element 1H of H is equal to theidentity element 1G of G.

Proposition 2.2 Let H be a non-empty subset of a group G. Then H is a subgroup of G,if and only if(i) h1, h2 ∈ H =⇒ h1h2 ∈ H; and(ii) h ∈ H =⇒ h−1 ∈ H.

Proof: H is a subgroup of G if and only if the four group axioms hold in H. Two ofthese, ‘Closure’, and ‘Inverses’, are the conditions (i) and (ii) of the lemma, and so if His a subgroup, then (i) and (ii) must certainly be true. Conversely, if (i) and (ii) hold,then we need to show that the other two axioms, ‘Associativity’ and ‘Identity’ hold in H.Associativity holds because it holds in G, and H is a subset of G. Since we are assumingthat H is non-empty, there exists h ∈ H, and then h−1 ∈ H by (ii), and hh−1 = 1 ∈ H by(i), and so ‘Identity’ holds, and H is a subgroup. 2

Examples 1. There are two standard subgroups of any group G: the whole group G itself,and the trivial subgroup {1} consisting of the identity alone. Subgroups other than G arecalled proper subgroups, and subgroups other than {1} are called non-trivial subgroups.

2. The non-zero real numbers R∗ form a subgroup of the multiplicative group of non-zerocomplex numbers C∗. Another subgroup of C∗ consists of the numbers z ∈ C with |z| = 1.

3. We have seen some examples in Subsection 1.3: SL(n,K) and O(n,K) are subgroups ofGL(n,K).

4. If g is any element of any group G, then the set of all powers {gx | x ∈ Z} forms asubgroup of G called the cyclic subgroup generated by g.

Let us look at a few specific examples. If G = (Z,+), then 5Z, which consists of all multiplesof 5, is the cyclic subgroup generated by 5. Of course, we can replace 5 by any integer here,but note that the cyclic groups generated by 5 and −5 are the same.

If G = 〈g〉 is a finite cyclic group of order n, and m is a positive integer dividing n, thenthe cyclic subgroup generated by gm has order n/m and consists of the elements gmk for0 ≤ k < n/m.

Exercise What is the order or the cyclic subgroup generated by gm for general m (where wedrop the assumption that m|n)?

In the dihedral groups (Subsection 1.10), the n rotations form a cyclic subgroup generatedby a.

Exercise Show that C∗ has finite cyclic subgroups of all possible orders.

5. (This is an example that was already introduces in Linear Algebra in connection withthe definition of the determinant.) Let G = Sym(X) be the symmetric group on the finiteset X (Subsection 1.4). We saw that every permutation in G can be written as a product ofdisjoint cycles. A cycle of length two is called a transposition. Since an arbitrary cycle like(1, 2, 3, . . . , n) can be written as a product of transpositions (for example, (1, 2, 3, . . . , n) =(1, 2)(2, 3)(3, 4) . . . (n− 1, n) ), we have:

Lemma 2.3 Any permutation on X can be written as a product of transpositions.

A permutation is called even if it is a product of an even number of transpositions, and oddif it is a product of an odd number of transpositions. In order for this to make much sense,we do need to prove the following:

11

Proposition 2.4 No permutation is both even and odd.

The proof of this proposition is at the end of Part 1 of the notes. It is not very instructive,and need not be memorised.

It is very easy to see that the set H of even permutations forms a subgroup of Sym(X). Thisis known as the alternating group on X, and is denoted by Alt(X). If (x, y) is some fixedtransposition on X then, for any g ∈ Sym(X), one of the two permutations g and g(x, y) iseven and the other is odd. Hence G = H ∪H(x, y), so |G : H| = 2.

As with Sym(X), the isomorphism type of Alt(X) depends only on |X|, and the notationAlt(n) or An is standard for the alternating group on a set X of size n.

Notice that a cycle of odd length is a product of an even number of transpositions, and acycle of even length is a product of an odd number of transpositions. This makes it easy todecide whether a given permutation in cyclic notation is even or odd (but it is also easy tomake mistakes!).

7. Let G = Sym(X) again, and let Y be a subset of a set X. Then the subset GY of Gdefined by GY = {g ∈ G | g(y) ∈ Y and g−1(y) ∈ Y for all y ∈ Y } is a subgroup of G.

Exercise If X is finite, what is the order of GY as a function of |Y | and |X|?

Definition A subgroup of Sym(X) for some set X is called a permutation group or a groupof permutations on X.

Lemma 2.5 The intersection of any set of subgroups of G is itself a subgroup of G.

2.2 Cosets and Lagrange’s Theorem

Throughout this subsection, H will be a subgroup of a group G.

Definition Let g ∈ G. Then the right coset Hg is the subset {hg | h ∈ H} of G. Similarly,the left coset gH is the subset {gh | h ∈ H} of G.

Note. In the case of additive groups, we denote the coset by H + g rather than by Hg.

Example Let G = D6 be the dihedral group of order 6 (Subsection 1.10). Then G consistsof the 6 permutations (), (1,2,3), (1,3,2), (1,2), (1,3), (2,3), where () represents the identitypermutation.

Let us first choose H = {(), (1, 2, 3), (1, 3, 2)} to be the cyclic subgroup generated by a =(1, 2, 3). If we put b = (2, 3), then we find that Hb = {(1, 2), (1, 3), (2, 3)}. In fact any rightcoset of H is equal to either H itself or to Hb = G \H. Furthermore, bH = Hb, and indeedHg = gH, for all g ∈ G, so the right and left cosets are the same in this example.

Now let us choose H = {(), (2, 3)} to be the cyclic subgroup generated by b = (2, 3).With a = (1, 2, 3), we have Ha = {(1, 2, 3), (1, 3)} and Ha2 = {(1, 3, 2), (1, 2)}, but aH ={(1, 2, 3), (1, 2)} and a2H = {(1, 3, 2), (1, 3)}, so the right and left cosets are not the same inthis case.

Note, however, that we always have g ∈ Hg and g ∈ gH.

Proposition 2.6 The following are equivalent for g, k ∈ G:(i) k ∈ Hg;(ii) Hg = Hk;(iii) kg−1 ∈ H.

12

Proof: Clearly Hg = Hk =⇒ k ∈ Hg, so (ii) =⇒ (i).

For (i) =⇒ (ii), suppose that k ∈ Hg. Then k = hg for some fixed h ∈ H. Multiplying thisequation on the left by h−1 gives g = h−1k. Let f ∈ Hg. Then, for some h1 ∈ H, we havef = h1g = h1h

−1k ∈ Hk, so Hg ⊆ Hk. Similarly, if f ∈ Hk, then for some h1 ∈ H, we havef = h1k = h1hg ∈ Hg, so Hk ⊆ Hg. Thus Hg = Hk, and we have proved that (i) =⇒ (ii).

If k ∈ Hg, then, as above, k = hg, and multiplying this on the right by g−1 gives kg−1 = h ∈H, so (i) =⇒ (iii).

Finally, if kg−1 ∈ H, then putting h = kg−1, we have hg = k, so k ∈ Hg, proving (iii) =⇒ (i).2

Corollary 2.7 Two right cosets Hg1 and Hg2 of H in G are either equal or disjoint.

Proof: If Hg1 and Hg2 are not disjoint, then there exists an element k ∈ Hg1 ∩Hg2, butthen Hg1 = Hk = Hg2 by the proposition. 2

Corollary 2.8 The right cosets of H in G partition G.

Proposition 2.9 If H is finite, then all right cosets have exactly |H| elements.

Proof: Since h1g = h2g =⇒ h1 = h2 by the cancellation law, it follows that the mapφ : H → Hg defined by φ(h) = hg is a bijection, and the result follows. 2

Of course, all of the above results apply with appropriate minor changes to left cosets.

Corollary 2.8 and Proposition 2.9 together imply:

Theorem 2.10 (Lagrange’s Theorem) Let G be a finite group and H a subgroup of G. Thenthe order of H divides the order of G.

Definition The number of distinct right cosets of H in G is called the index of H in G andis written as |G : H|.If G is finite, then we clearly have |G : H| = |G|/|H|.Exercise Prove that, if |G : H| is finite, then |G : H| is also equal to the number of distinctleft cosets of H in G. This is clear if G is finite, because both numbers are equal to |G|/|H|,but it is not quite so easy if G is infinite.

Proposition 2.11 Let G be a finite group. Then for any g ∈ G, the order |g| of g dividesthe order |G| of G.

Proof: Let |g| = n. We saw in Example 4 of Subsection 2.1 that the powers {gx | x ∈ Z}of g form a subgroup H of G, and we saw in Subsection 1.6 that the distinct powers of g are{gx | 0 ≤ x < n}. Hence |H| = n and the result follows from Lagrange’s Theorem. 2

As an application, we can now immediately classify all finite groups whose order is prime.

Proposition 2.12 Let G be a group having prime order p. Then G is cyclic; that is, G ∼= Cp.

Proof: Let g ∈ G with 1 6= g. Then |g| > 1, but |g| divides p by Proposition 2.11, so |g| = p.But then G must consist entirely of the powers gk (0 ≤ k < p) of g, so G is cyclic. 2

This result and Proposition 1.11 provide a classification of all groups of order less than 6.We shall soon (Subsection 3.2) be able to deal with groups of order 6 too.

13

3 Normal Subgroups and Quotient Groups

3.1 Normal Subgroups

Definition A subgroup H of a group G is called normal in G if the left and right cosets gHand Hg are equal for all g ∈ G.

The standard notation for “H is a normal subgroup of G” is H � G or H � G. (H � Gis sometimes but not always used to mean that H is a proper normal subgroup of G – i.e.H 6= G.)

Examples 1. The two standard subgroups G and {1} of any group G are both normal.

2. Any subgroup of an abelian group is normal.

3. In the exampleG = D6 in Subsection 2.2, we saw that the subgroup {(), (1, 2, 3), (1, 3, 2)} (={1, a, a2}) is normal in G, but the subgroup {(), (2, 3)} = {1, b} is not normal in G.

In Example 3, the normal subgroup H = {1, a, a2} has index |G|/|H| = 6/3 = 2 in G. In factwe have the general result:

Proposition 3.1 If G is any group and H is a subgroup with |G : H| = 2, then H is anormal subgroup of G.

Proof: Assume that |G : H| = 2. Then there are only two distinct right cosets of G, one ofwhich is H, and so by Corollary 2.8, the other one must be G \H. The same applies to leftcosets. Hence, for g ∈ G, if g ∈ H then gH = Hg = H and if g 6∈ H then gH = Hg = G \H.In either case gH = Hg, so H �G. 2

So, in Example 6 of Subsection 2.1, Alt(X) is a normal subgroup of Sym(X).

The following result often provides a useful method of testing a subgroup for normality.

Proposition 3.2 Let H be a subgroup of the group G. Then H is normal in G if and onlyif ghg−1 ∈ H for all g ∈ G and h ∈ H.

Proof: Suppose that H � G, and let g ∈ G, h ∈ H. Then Hg = gH, and gh ∈ gH, sogh ∈ Hg, which means that there exists h′ ∈ H with gh = h′g. Hence ghg−1 = h′ ∈ H.

Conversely, assume that ghg−1 ∈ H for all g ∈ G, h ∈ H. Then for gh ∈ gH, we haveghg−1 ∈ H, so gh = h′g for some h′ ∈ H; i.e. gh ∈ Hg, and we have shown that gH ⊆ Hg.For hg ∈ Hg, we have g−1hg ∈ H (because g−1hg = ghg−1 where g = g−1), so, puttingh′ = g−1hg, we have hg = gh′ ∈ gH, and so Hg ⊆ gH. Thus gH = Hg, and H �G. 2

Example 4. Let G be the dihedral group D12. Recall from Subsection 1.10 that G ={1, a, a2, a3, a4, a5, b, ab, a2b, a3b, a4b, a5b} where a and b are the permutations (1, 2, 3, 4, 5, 6)and (2, 6)(3, 4); the full multiplication table of this group is written out in Subsection 1.10.

Let H be the cyclic subgroup {1, a3} of order 2 of G. We claim that H is normal in G.To prove this, we have to show that ghg−1 ∈ H for all g ∈ G, h ∈ H. If h = 1, thenghg−1 = gg−1 = 1 ∈ H, so we only need consider h = a3. Then if g = ak for 0 ≤ k ≤ 5,we have ghg−1 = aka3a−k = a3 ∈ H. The remaining case is g = akb for 0 ≤ k ≤ 5.From Subsection 1.10, we have ba3 = a6−3b = a3b and so ba3b−1 = a3, and then, for any k,(akb)a3(akb)−1 = akba3b−1a−k = aka3a−k = a3 ∈ H, and we have proved that H �G.

3.2 Classification of Groups of Orders 6 and 8

The following easy lemma was on an Assignment Sheet:

14

Lemma 3.3 Let G be a group in which g2 = 1 for all g ∈ G. Then G is abelian.

Lemma 3.4 Let G be a group in which g2 = 1 for all g ∈ G, and let a, b be distinct non-identity elements of G. Then {1, a, b, ab} is a subgroup of G of order 4.

Proof: We cannot have ab = 1, a or b (exercise), so |{1, a, b, ab}| = 4. To prove it is asubgroup, we check that (i) and (ii) of Proposition 2.2 hold. Since g2 = 1 for all g ∈ G, wehave g = g−1 for all g ∈ G, and so (ii) is satisfied. (i) is straightforward, given that G isabelian and a2 = b2 = 1. For example, b(ab) = bab = bba = a. 2

We can now classify groups of order 6. We have seen two examples so far, the cyclic groupC6 and the dihedral group D6. (Notice that the symmetric group S3 is actually equal toD6.) Since D6 is non-abelian (or, alternatively, since it has no element of order 6), thesetwo groups cannot be isomorphic to each other. We shall now prove that they are the onlyexamples.

Proposition 3.5 Let G be a group of order 6. Then G ∼= C6 or G ∼= D6.

Proof: By Proposition 2.11 the orders of elements g ∈ G can be 1, 2, 3 or 6. If there is a gwith |g| = 6, then G ∼= C6, so assume not. If all elements had order 1 or 2, then by the lastlemma, G would have a subgroup with four elements, which contradicts Lagrange’s Theorem.Hence there is an element a of order 3. Then N = {1, a, a2} has index 2 in G, so it is a normalsubgroup by Proposition 3.1. Choose b ∈ G \N . Then G = N ∪Nb = {1, a, a2, b, ab, a2b}.What can b2 be? b2 = b, ab or a2b all lead to contradictions by the cancellation law. If b2 = aor a2 on the other hand, then b has order 6, contrary to assumption. So b2 = 1.

By Proposition 3.2, we have bab−1 ∈ N . If bab−1 = 1, then ba = b which is false. If bab−1 = a,then ba = ab and then we find that ab has order 6, contrary to assumption. The remainingpossibility is bab−1 = a2, and then ba = a2b = a−1b and G ∼= D6 by Proposition 1.12. 2

The classification of groups of order 8 takes a little more effort, because there are five iso-morphism classes, but it can be done using only the results proved so far.

Proposition 3.6 Let G be a group of order 8. Then G is isomorphic to one of C8, C4×C2,C2 × C2 × C2, D8 and Q8.

Proof: If G has an element of order 8, then it is cyclic (C8), so assume not. If all non-identity elements have order 2, then G is abelian by Lemma 3.3, and by Lemma 3.4 there isa subgroup H = {1, a, b, ab} isomorphic to a Klein Four Group C2 × C2. Choose c ∈ G \H.Then G = H ∪Hc, and it can be checked that G ∼= H × 〈c〉 ∼= C2 × C2 × C2.

Otherwise, there is an element a ∈ G of order 4, and a normal subgroup N = {1, a, a2, a3}.Let b ∈ G \N , so G = N ∪Nb. As in Proposition 3.5, we cannot have b2 ∈ Nb, so b2 ∈ N .If b2 = a or a3, then |b| = 8, contrary to assumption. Hence b2 = 1 or a2. Since N isnormal, bab−1 ∈ N . As in Proposition 3.5, we cannot have bab−1 = 1. If bab−1 = a2, thenba2b−1 = bab−1bab−1 = a2a2 = 1, and then a2 = b−1b = 1, contradiction, so bab−1 6= a2, andhence bab−1 = a or a3; that is, ba = ab or ba = a3b = a−1b. We now have four possibilitiesto analyse:(i) b2 = 1, ba = ab. Then G ∼= C4 × C2 by Proposition 1.13.(ii) b2 = a2, ba = ab. In this case, we have (ab)2 = a2b2 = a4 = 1, so if we replace b by ab,

then we are back in Case (i), and G ∼= C4 × C2 again.(iii) b2 = 1, ba = a−1b. G ∼= D8 by Proposition 1.12.(iv) b2 = a2, ba = a−1b. G ∼= Q8 by Proposition 1.14.

2

15

3.3 Quotient Groups

The definition of quotient groups depends on the following crucial technical result.

Lemma 3.7 Let N be a normal subgroup of a group G, and let g, h ∈ G. Then the productof any element in the coset Ng with any element in the coset Nh is equal to an element inthe coset Ngh.

Proof: Let n1g ∈ Ng and n2h ∈ Nh. Then, by normality of N , we have gN = Ng, andso gn2 is equal to some element n3g ∈ Ng. Hence (n1g)(n2h) = n1(gn2)h = n1(n3g)h =(n1n3)gh ∈ Ngh, which proves the lemma. 2

Definition If A and B are subsets of a group G, then we define their product AB = {ab |a ∈ A, b ∈ B}.

Lemma 3.8 If N is a normal subgroup of G and Ng, Nh are cosets of N in G, then(Ng)(Nh) = Ngh.

Proof: By Lemma 3.7, we have (Ng)(Nh) ⊆ Ngh, but ngh = (ng)(1h) ∈ (Ng)(Nh), soNgh ⊆ (Ng)(Nh), and we have equality. 2

Theorem 3.9 Let N be a normal subgroup of a group G. Then the set G/N of right cosetsNg of N in G forms a group under multiplication of sets.

Proof: We have just seen that (Ng)(Nh) = Ngh, so we have closure, and associativityfollows easily from associativity of G. Since (N1)(Ng) = N1g = Ng for all g ∈ G, N1 is anidentity element, and since (Ng−1)(Ng) = Ng−1g = N1, Ng−1 is an inverse to Ng for allcosets Ng. Thus the four group axioms are satisfied and G/N is a group. 2

Definition The group G/N is called the quotient group (or the factor group) of G by N .

Notice that if G is finite, then |G/N | = |G : N | = |G|/|N |.So, although the quotient group seems a rather complicated object at first sight, it is actuallya smaller group than G. If we can find a normal subgroup N of a group G, then we havepartially succeeded in splitting up the study of G into the study of the two smaller groups Nand G/N . Groups that cannot be split up in this sense are called simple. More precisely:

Definition A groupG with |G| > 1 is called simple if its only normal subgroups areG and {1}.

Example Cyclic groups of prime order p are simple, because by Lagrange’s Theorem, theonly possible orders of their subgroups are 1 and p.

Proposition 3.10 A simple abelian group is cyclic of prime order.

Proof: All subgroups of an abelian group G are normal, so we have only to find a propernon-trivial subgroup to establish non-simplicity. Let G be a simple abelian group and choose1 6= g ∈ G. If |g| is infinite, then the subgroup generated by g2 is non-trivial and proper, soG is not simple. If |g| is finite but not prime, say |g| = lm, then the subgroup generated bygl is proper and non-trivial, so G is not simple. Hence |g| = p is prime, and we must have〈g〉 = G, or 〈g〉 would be a proper non-trivial subgroup. Hence G is cyclic of prime order. 2

3.4 Examples of Quotient Groups

1. Let G be the infinite cyclic group (Z,+), and let N = nZ be its subgroup generated by afixed positive integer n – let’s take n = 5 just to be specific. Now, by using Lemma 2.6 (after

16

changing from multiplicative to additive notation!), we see that the cosets 5Z+k and 5Z+j ofN in G are equal if and only if k ≡ j (mod 5). So there are only 5 distinct cosets, namelyN = N+0, N+1, N+2, N+3, N+4. It is now clear that G/N is isomorphic to the groupZ5 = {0, 1, 2, 3, 4} (see Subsection 1.7) under the correspondence N+i→ i for 0 ≤ i < 5.

2. Now let G = 〈g〉 be finite cyclic, and suppose that |g| = lm is composite. Let N be thenormal subgroup 〈gm〉. We saw in Example 4 of Subsection 2.1 thatN has order l and consistsof the elements {gmk | 0 ≤ k < l}. Since all cosets have the form Ngk for some k ∈ Z, it isclear that G/N is cyclic and is generated by Ng. We can calculate its order as |G|/|N | = m.To see this directly, note that (using Lemma 2.6 again) Ngk = Ngj ⇔ gk−j ∈ N ⇔ m|(k−j),and so the distinct cosets are Ngk for 0 ≤ k < m. In particular, (Ng)m = Ngm = N1 is theidentity element of G/N , and |Ng| = m.

3. For a more complicated example, we take Example 4 of Subsection 3.1, namely G = D12

and N = {1, a3}. Then |G/N | = |G|/|N | = 6. For g ∈ G, let us denote Ng by g. (This is acommonly used notation, but you must always keep in mind that g = h does not necessarilyimply that g = h!) Then, since a3 ∈ N , we have

a3 = (Na)3 = Na3 = N1 = 1

is the identity of G/N . We also have b2 = 1 and ba = a−1b, because these relations areinherited from the corresponding relations of G. Thus G/N is a group of order 6 satisfyingthe three relations a3 = 1, b2 = 1, ba = a−1b, and so by Proposition 1.12 G/N ∼= D6.

It might be helpful in understanding this example to see the full multiplication table ofG again (we saw it already in Subsection 1.10), but this time with the elements arrangedaccording to their cosets. Notice that all elements in each 2× 2 block of this table lie in thesame coset of N in G. We can then see the multiplication table of G/N by regarding these2× 2 blocks as single elements (i.e. cosets) in the quotient group.

N Na Na2 Nb Nab Na2b1 a3 a a4 a2 a5 b a3b ab a4b a2b a5b

N 1 1 a3 a a4 a2 a5 b a3b ab a4b a2b a5ba3 a3 1 a4 a a5 a2 a3b b a4b ab a5b a2b

Na a a a4 a2 a5 a3 1 ab a4b a2b a5b a3b ba4 a4 a a5 a2 1 a3 a4b ab a5b a2b b a3b

Na2 a2 a2 a5 a3 1 a4 a a2b a5b a3b b a4b aba5 a5 a2 1 a3 a a4 a5b a2b b a3b ab a4b

Nb b b a3b a5b a2b a4b ab 1 a3 a5 a2 a4 aa3b a3b b a2b a5b ab a4b a3 1 a2 a5 a a4

Nab ab ab a4b b a3b a5b a2b a a4 1 a3 a5 a2

a4b a4b ab a3b b a2b a5b a4 a a3 1 a2 a5

Na2b a2b a2b a5b ab a4b b a3b a2 a5 a a4 1 a3

a5b a5b a2b a4b ab a3b b a5 a2 a4 a a3 1

4 Homomorphisms and the Isomorphism Theorems

4.1 Homomorphisms

Definition Let G and H be groups. A homomorphism φ from G to H is a map φ : G→ Hsuch that φ(g1g2) = φ(g1)φ(g2) for all g1, g2 ∈ G.

A homomorphism φ is called a monomorphism if it is an injection; that is, if φ(g1) = φ(g2) =⇒g1 = g2.

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A homomorphism φ is called an epimorphism if it is a surjection; that is, if im(φ) = H.

A homomorphism φ is called an isomorphism if it is a bijection; that is, if it is both amonomorphism and an epimorphism.

Lemma 4.1 Let φ : G → H be a homomorphism. Then φ(1G) = 1H and φ(g−1) = φ(g)−1

for all g ∈ G.

Proof: (Recall that 1G and 1H are the identity elements of G and H.) Let φ(1G) = h. Then

1Hh = h = φ(1G) = φ(1G1G) = φ(1G)φ(1G) = hh,

so h = 1H by the cancellation law. Similarly, if g ∈ G and φ(g) = h, then

φ(g−1)φ(g) = φ(g−1g) = φ(1G) = 1H = h−1h = φ(g)−1φ(g)

so φ(g−1) = φ(g)−1 by the cancellation law. 2

Examples 1. If H is a subgroup of G, then the map φ : H → G defined by φ(h) = h for allh ∈ H is a monomorphism. It is an isomorphism if H = G.

2. If G is an abelian group and r ∈ Z, then (gh)r = grhr for all g, h ∈ G, so the mapφ : G→ G defined by φ(g) = gr is a homomorphism.Warning. This only works when G is abelian.

3. Let k be a fixed element of a group G. Then, for g, h ∈ G, we have kghk−1 = kgk−1khk−1,so the map φ : G → G defined by φ(g) = kgk−1 is a homomorphism. In fact it is anisomorphism, because kgk−1 = khk−1 =⇒ g = h by the cancellation laws, and each h ∈ Gis equal to φ(k−1hk). Notice that if G is abelian, then whatever k we choose, we always getφ(g) = g for all g, so these examples are only interesting for non-abelian groups.

We can deduce the following result from Proposition 1.10.

Proposition 4.2 If g, k are elements of a group G, then g and kgk−1 have the same order.

The elements g and kgk−1 are called conjugate elements. We shall be studying this relation-ship further in Subsection 5.3.

4. Let G = {1, a, b, c} be a Klein Four Group. Define φ : G → G by φ(1) = 1, φ(a) =b, φ(b) = c, φ(c) = a. Then it is straightforward to check that φ is an isomorphism.

Exercise Find lots more homomorphisms from G to G in this case, including some that arenot isomorphisms.

5. Recall that GL(n,K) is the group of invertible n× n matrices over the field K, and thatK∗ is the multiplicative group of non-zero elements of K. By the product of determinantsrule, det(AB) = det(A) det(B), it follows that the map φ : GL(n,K) → K∗ defined byφ(g) = det(g) is a homomorphism.

6. Here is an example of a homomorphism from an additive group to a multiplicative group.Let G = (R,+) and H = R∗, and define φ : G → H by φ(g) = exp(g). Then φ(g1 + g2) =φ(g1)φ(g2), which says that φ is a homomorphism. In fact φ is a monomorphism but not anepimorphism, because im(φ) is the subgroup of positive real numbers.

4.2 Kernels and Images

Definition Let φ : G → H be a homomorphism. Then the kernel ker(φ) of φ is defined tobe the set of elements of G that map onto 1H ; that is,

ker(φ) = { g ∈ G | φ(g) = 1H }.

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Note that by Lemma 4.1 above, ker(φ) always contains 1G.

Proposition 4.3 Let φ : G → H be a homomorphism. Then φ is a monomorphism if andonly if ker(φ) = {1G}.

Proof: Since 1G ∈ ker(φ), if φ is a monomorphism, then we must have ker(φ) = {1G}.Conversely, suppose that ker(φ) = {1G}, and let g1, g2 ∈ G with φ(g1) = φ(g2). Then1H = φ(g1)−1φ(g2) = φ(g−1

1 g2) (by Lemma 4.1), so g−11 g2 ∈ ker(φ) and hence g−1

1 g2 = 1G

and g1 = g2. So φ is a monomorphism. 2

The following theorem says that the set of normal subgroups of G is equal to the set of kernelsof homomorphisms with domain G.

Theorem 4.4 (i) Let φ : G→ H be a homomorphism. Then ker(φ) is a normal subgroup of G.(ii) Let N be a normal subgroup of a group G. Then the map φ : G → G/N defined byφ(g) = Ng is a homomorphism (in fact an epimorphism) with kernel N .

Proof: (i) Checking that ker(φ) is a subgroup of G is straightforward, using Proposition 2.2.If g ∈ G and k ∈ K, then

φ(gkg−1) = φ(g)φ(k)φ(g−1) = φ(g)1Kφ(g−1) = 1K ,

so gkg−1 ∈ ker(φ) and K �G by Proposition 3.2.(ii) It is straightforward to check that φ is an epimorphism, and φ(g) = 1H ⇔ Ng = N1G ⇔g ∈ N , so ker(φ) = N . 2

The epimorphism defined in (ii) of the above theorem is called the natural or canonicalhomomorphism/epimorphism from G to G/N .

The image im(φ) of a homomorphism is just its image as a map, and the following propositionis straightforward to prove.

Proposition 4.5 Let φ : G→ H be a homomorphism. Then im(φ) is a subgroup of H (butnot necessarily a normal subgroup).

Example 7. Let G = H = D12, the dihedral group of order 12. We saw in Subsection 1.10that G = { ak | 0 ≤ k < 6 } ∪ { akb | 0 ≤ k < 6 }. We define φ : G → H by φ(ak) = a2k andφ(akb) = a2kb for 0 ≤ k < 6. We claim that φ is a homomorphism. It seems at first sight asthough we need to check that φ(gh) = φ(g)φ(h) for all 144 ordered pairs g, h ∈ G, but wecan group these tests into the four distinct types listed in Subsection 1.10. We will make freeuse of the fact that am = 1 when 6|m.(i) φ(akal) = φ(ak+l) or φ(ak+l−6) = a2(k+l) or a2(k+l−6) = a2ka2l = φ(ak)φ(al);(ii) φ(ak(alb)) = φ(ak)φ(alb) – this is similar to (i);(iii) φ((akb)al) = φ(ak−lb) or φ(ak−l+6b) = a2(k−l)b or a2(k−l+6)b = a2ka−2lb = a2kba2l =

φ(akb)φ(al)(iv) φ((akb)(alb)) = φ(akb)φ(alb) – this is similar to (iii).

So φ really is a homomorphism. We can check that the only elements of G with φ(g) = 1 areg = 1 and g = a3, so ker(φ) = {1, a3}, which is the normal subgroup that we considered inExample 3 of Subsection 3.4. im(φ) consists of the 6 elements 1, a2, a4, b, a2b, a4b of G.

In general, if φ : G → H is a homomorphism and J is a subset of H, then we define thecomplete inverse image of J under φ to be the set φ−1(J) = { g ∈ G | φ(g) ∈ J }. It is easyto check, using Proposition 2.2, that if J is a subgroup of H, then φ−1(J) is a subgroup ofG.

19

In the case where φ is the canonical epimorphism G → G/N of Theorem 4.4(ii), φ−1(J) isa union of cosets of N , and if J is a subgroup, then φ−1(J) is a subgroup of G containingN . Conversely, if I is a subgroup of G containing N , then J = I/N is a subgroup of G/Nwith φ−1(J) = I, and so we get the following proposition, which is useful when working withquotient groups. We shall use it in the proof of Sylow’s Theorem (5.14) later on.

Proposition 4.6 If N � G, then the subgroups of G/N are precisely the quotient groupsI/N , for subgroups I of G that contain N .

4.3 The Isomorphism Theorems

Theorem 4.7 (First Isomorphism Theorem) Let φ : G→ H be a homomorphism with kernelK. Then G/K ∼= im(φ). More precisely, there is an isomorphism φ : G/K → im(φ) definedby φ(Kg) = φ(g) for all g ∈ G.

Proof: The trickiest point to understand in this proof is that we have to show that φ(Kg) =φ(g) really does define a map from G/K to im(φ). The reason that this is not obvious isthat we can have Kg = Kh with g 6= h, and when that happens we need to be sure thatφ(g) = φ(h). This is called checking that the map φ is well-defined. In fact, once you haveunderstood what needs to be checked, then doing it is quite easy, because Kg = Kh =⇒ g =kh for some k ∈ K = ker(φ), and then φ(g) = φ(k)φ(h) = φ(h).

Clearly im(φ) = im(φ), and it is straightforward to check that φ is a homomorphism. Finally,

φ(Kg) = 1H ⇐⇒ φ(g) = 1H ⇐⇒ g ∈ K ⇐⇒ Kg = K1 = 1G/K ,

and so φ is a monomorphism by Proposition 4.3. Thus φ : G/K → im(φ) is an isomorphism,which completes the proof. 2

Let us illustrate this theorem using Example 6 from the last subsection. Note that theelements of G = D12 are listed in two separate columns in the diagram, in different orders,once for the domain and once for the codomain of φ. The elements of imφ are printed slightlyto the left of those not in im(φ) in the codomain column.

φ

φ

1= N

a3 } - 1

a= Na

a

a4 } - a2

a2

= Na2

a3

a5 } - a4

b= Nb

a5

a3b} - b

ab= Nab

ab

a4b} - a2b

a2b= Na2b

a3b

a5b} - a4b

a5b

The other two isomorphism theorems are less important, and are used mainly in more ad-vanced courses on group theory. Before we can state the Second Isomorphism Theorem, we

20

need a lemma. Recall from Subsection 3.7 that the product of two subsets A and B of G isdefined as AB = { ab | a ∈ A, b ∈ B }. This is not usually a subgroup of G, even when A andB are subgroups (can you find an example to demonstrate this?). However, we have:

Lemma 4.8 If H is any subgroup and K is a normal subgroup of a group G, then HK = KHis a subgroup of G.

Proof: Let hk ∈ HK. Then, by normality of K, hk ∈ hK = Kh ⊆ KH so HK ⊆ KHand similarly KH ⊆ HK, and we have equality. The product of two elements of HK liesin HKHK = HHKK = HK, and the inverse k−1h−1 of an element hk ∈ HK lies inKH = HK, so HK is a subgroup of G by Proposition 2.2. 2

Theorem 4.9 (Second Isomorphism Theorem) Let H be any subgroup and let K be a normalsubgroup of a group G. Then H ∩K is a normal subgroup of H and H/(H ∩K) ∼= HK/K.

Proof: Use Proposition 3.2 to show that H ∩K�H. Let φ : G→ G/K be the natural map(see Theorem 4.4(ii)). Then φ(H) is the set of cosets Kh for h ∈ H, which together formthe subgroup KH/K = HK/K of G/K; in other words im(φH) = HK/K. Also ker(φH) =H ∩ker(φ) = H ∩K. Now, by applying Theorem 4.7 to φH , we get H/(H ∩K) ∼= HK/K. 2

Example Let K and H be respectively the subgroups {1, a2, a4} and {1, a3, b, a3b} of G =D12. Then K ∼= C3 and H is a Klein Four Group. K is a normal subgroup: to check this,use Proposition 3.2; for example,

(akb)a2(akb)−1 = akba2b−1a−k = aka4a−k = a4

for all k ∈ Z. Clearly H ∩K = {1} is trivial, and so H/(H ∩K) ∼= H is also a Klein FourGroup. By direct calculation, we find HK = G, so HK/K = G/K is a Klein Four Group.

Theorem 4.10 (Third Isomorphism Theorem) Let K ⊆ H ⊆ G, where H and K are bothnormal subgroups of G. Then (G/K)/(H/K) ∼= G/H.

Proof: Define φ : G/K → G/H by φ(Kg) = Hg for all g ∈ G. As in the proof ofTheorem 4.7, we need to check that this is well-defined; that is, that Kg1 = Kg2 =⇒ φ(g1) =φ(g2). This is easy, because K ⊂ H, so Kg1 = Kg2 =⇒ Hg1 = Hg2.

Since im(φ) = G/H and ker(φ) = H/K, the result follows by applying Theorem 4.7 to φ. 2

5 Groups Acting On Sets

5.1 Definition and Examples

Definition Let G be a group and X a set. An action of G on X is a map · : G ×X → X,which satisfies the properties:(i) (gh) · x = g · (h · x) for all g, h ∈ G, x ∈ X;(ii) 1G · x = x for all x ∈ X.

Note 1. In this definition, the image of (g, x) under the map · is denoted by g · x.2. We have actually defined a left action of G on X. A right action can be defined analogouslyas a map X ×G→ X.

Proposition 5.1 Let · be an action of the group G on the set X. For g ∈ G, define themap φ(g) : X → X by φ(g)(x) = g · x. Then φ(g) ∈ Sym(X), and φ : G → Sym(X) is ahomomorphism.

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Proof: Property (ii) in the definition says that φ(1G) is the identity map IX : X → X, andthen Property (i) implies that φ(g)φ(g−1) = φ(gg−1) = IX , and similarly φ(g−1)φ(g) = IX .So φ(g) and φ(g−1) are inverse maps, which proves that φ(g) : X → X is a bijection. Henceφ(g) ∈ Sym(X), and then Property (i) implies immediately that φ is a homomorphism. 2

The kernel of an action · of G on X is defined to be the kernel K = ker(φ) of the homomor-phism φ : G→ Sym(X) defined in Proposition 5.1. So

K = { g ∈ G | g · x = x for all x ∈ X }.

The action is said to be faithful if K = {1}. In this case, Theorem 4.7 says that G ∼= G/K ∼=im(φ), which we state as a proposition:

Proposition 5.2 If · is a faithful action of G on X, then G is isomorphic to a subgroup ofSym(X).

Examples 1. If G is a subgroup of Sym(X), then we can define an action of G on X simplyby putting g · x = g(x) for x ∈ X. This action is faithful.

2. Let P be a regular hexagon, and let G = {1, a, a2, a3, a4, a5, b, ab, a2b, a3b, a4b, a5b} = D12

be the group of isometries of P . In Subsection 1.10, we defined a and b to be the permutations(1, 2, 3, 4, 5, 6) and (2, 6)(3, 5) of the set {1, 2, 3, 4, 5, 6} of vertices of P , and so this immediatelygives us an action of G on the vertex set.

There are some other related actions however. We could instead take X to be the set E ={e1, e2, e3, e4, e5, e6} of edges of P , where e1 is the edge joining 1 and 2, e2 joins 2 and3, etc. The map φ of the action of G on E is then given by φ(a) = (e1, e2, e3, e4, e5, e6),φ(b) = (e1, e6)(e2, e5)(e3, e4). (Notice that any homomorphism is fully specified by the imagesof a set of group generators, because the images of all other elements in the group can becalculated from these.) This action is still faithful.

As a third possibility, let D = {d1, d2, d3} be the set of diagonals of P , where d1 joins vertices1 and 4, d2 joins 2 and 5, and d3 joins 3 and 6. Then map φ of the action of G on D is definedby φ(a) = (d1, d2, d3), and φ(b) = (d2, d3). This action is not faithful, and its kernel is thenormal subgroup {1, a3} of G that we have already studied. The image is isomorphic to D6.

3. There is a faithful action called the left regular action, which we can define for any groupG. Here we put X to be the underlying set of G and simply define g · x to be gx for allg ∈ G, x ∈ X. Conditions (i) and (ii) of the definition obviously hold, so we have defined anaction. If g is in the kernel K of the action, then gx = x for all x ∈ X, which implies g = 1by the cancellation law, so the action is faithful. From Proposition 5.2, we can now deduce:

Theorem 5.3 (Cayley’s Theorem) Every group G is isomorphic to a permutation group.(That is, to a subgroup of Sym(X) for some set X.)

5.2 Orbits and Stabilisers

Definition Let · be an action of G act on X. We define a relation ∼ on X by x ∼ y if andonly if there exists a g ∈ G with y = g ·x. Then ∼ is an equivalence relation – the proof is leftas an exercise. The equivalence classes of ∼ are called the orbits of G on X. In particular,the orbit of a specific element x ∈ X, which is denoted by G · x or by OrbG(x) is

{ y ∈ X | ∃g ∈ G with g · x = y }.

In most of the examples that we have seen so far, there is just one orbit, the whole of X.There was one example with two orbits. Let X be a set, and Y a proper non-empty subset

22

of X. In Example 7 of Subsection 2.1, we defined the subgroup Sym(X)Y of Sym(X) to be{ g | g ∈ Sym(X), g(y) ∈ Y and g−1(y) ∈ Y ∀y ∈ Y }. Denote this subgroup by G, and giveit the obvious action on X with g · x = g(x) for g ∈ G, x ∈ X. Then there are two orbits, Yand X \ Y .

In a similar way, for any partition of X, we can can define a subgroup of Sym(X) having thesets in this partition as the orbits.

Definition Let G act on X and let x ∈ X. Then the stabiliser of x in G, which is denotedby Gx or by StabG(x) is { g ∈ G | g · x = x }.The proof of the following proposition is left as an exercise.

Proposition 5.4 Let G act on X and x ∈ X. Then(i) StabG(x) is a subgroup of G;(ii) ∩x∈XStabG(x) is the kernel of the action of G on X.

The next theorem is a very fundamental result in group theory.

Theorem 5.5 (The Orbit-Stabiliser Theorem) Let the finite group G act on X, and letx ∈ X. Then |G| = |OrbG(x)| |StabG(x)|.

Proof: Let y ∈ OrbG(x). Then there exists a g ∈ G with g · x = y. Let H = StabG(x). Foran element g′ ∈ G, we have

g′ · x = y ⇐⇒ g′ · x = g · x⇐⇒ g−1g′ · x = x⇐⇒ g−1g′ ∈ H ⇐⇒ g′ ∈ gH.

So the elements g′ with g′ · x = y are precisely the elements of the coset gH. But, byProposition 2.9 (applied to left rather than to right cosets!), we have |gH| = |H|. In otherwords, for each y ∈ OrbG(x), there are precisely |H| elements g′ of G with g′ · x = y. Hencethe total number of such y ∈ OrbG(x) must be |G|/|H|, which proves the result. 2

5.3 Conjugacy Classes

In Example 3 of Subsection 5.1, a group G was made to act on the set of its own elementsby multiplication on the left; that is, g · x = gx for g, x ∈ G.

There is another important action of G on X = G, which is defined by

g · x = gxg−1 for g, x ∈ G.

It is easy to check that conditions (i) and (ii) of the definition hold, so this does indeed definean action. This action is called conjugation. The orbits of the action are called the conjugacyclasses of G, and elements in the same conjugacy class are said to be conjugate in G. Sog, h ∈ G are conjugate if and only if there exists f ∈ G with h = fgf−1. We will writeClG(g) for the orbit of g; that is the conjugacy class containing g. We have seen already inProposition 4.2 that conjugate elements have the same order.

What is StabG(g) for this action? By definition it consists of the elements f ∈ G for whichf · g = g; that is, fgf−1 = g, or equivalently fg = gf . In other words, it consists of those fthat commute with g. It is called the centraliser of g in G and is written as CG(g).

By applying the Orbit-Stabiliser Theorem (5.5), we get:

Proposition 5.6 Let g ∈ G. Then |ClG(g)| = |G|/|CG(g)|.

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The kernel K of the action consists of those f ∈ G that fix and hence commute with allg ∈ G. This is called the centre of G and is denoted by Z(G). So we have

Z(G) = { f ∈ G | fg = gf ∀g ∈ G }.Note that g ∈ Z(G) if and only if ClG(g) = {g}.It is high time that we worked out some examples!

Examples 1. Let G be an abelian group. Then Z(G) = G, CG(g) = G and ClG(g) = {g}for all g ∈ G.

2. Let G = { ak | 0 ≤ k < n } ∪ { akb | 0 ≤ k < n } be the dihedral group D2n. Then ofcourse, as always, ClG(1) = {1}. To compute the remaining classes, we calculate fgf−1 inthe four cases, when f and g have the form ak (1 ≤ k < n) and akb (0 ≤ k < n). Sinceb2 = 1, we have b = b−1, so we shall always substitute b for b−1 in our calculations.

(i) f = al, g = ak: fgf−1 = g.(ii) f = alb, g = ak: fgf−1 = bakb = a−k = g−1.(iii) f = al, g = akb: fgf−1 = ak+lba−l = ak+lalb = ak+2lb.(iv) f = alb, g = akb: fgf−1 = albakbba−l = albak−l = alal−kb = a2l−kb.

The cases when n is odd and even are different. Suppose first that n is odd. Then, by (i)and (ii), a and a−k = an−k are conjugate for all k, and we have the distinct conjugacy classes{ak, an−k} for 1 ≤ k ≤ (n− 1)/2, all of which contain just two elements. By (iii), we see thatb is conjugate to a2lb for 0 ≤ l < n, and when n is odd, this actually includes all elementsalb for 0 ≤ l < n. (For example, ab = a2lb with l = (n+ 1)/2.) So the set { akb | 0 ≤ k < n }forms a single conjugacy class. Geometrically, this is not surprising, because these n elementsare all reflections that pass through one vertex and the centre of the polygon P of which Gis the group of isometries.

Now suppose that n is even. Then, when k = n/2, we have ak = a−k, and so {an/2} is aconjugacy class of size 1 (and hence an/2 ∈ Z(G)). We also have the classes {ak, an−k} of size2 for 1 ≤ k ≤ (n − 2)/2. In this case, the reflections akb split up into two conjugacy classesof size n/2, namely { a2kb | 0 ≤ k < n/2 } and { a2k+1b | 0 ≤ k < n/2 }. Geometrically thesecorrespond to the two different types of reflections: those about lines that pass through twovertices of P and those about lines that bisect two edges of P .

3. Let G = Sym(X) and let f, g ∈ G. Let us write g in cyclic notation, and suppose thatone of the cycles of g is (x1, x2, . . . , xr). Then g(x1) = x2, and so fg(x1) = f(x2) andhence fgf−1(f(x1)) = f(x2). Similarly, we have fgf−1(f(xi)) = f(xi+1) for 1 ≤ i < r andfgf−1(f(xr)) = f(x1). Hencefgf−1 has a cycle (f(x1), f(x2), . . . , f(xr)), and we have:

Proposition 5.7 Given a permutation g in cyclic notation, we obtain the conjugate fgf−1

of g by replacing each element x ∈ X in the cycles of g by f(x).

For example, if X = {1, 2, 3, 4, 5, 6, 7}, g = (1, 5)(2, 4, 7, 6) and f = (1, 3, 5, 7, 2, 4, 6), thenfgf−1 = (3, 7)(4, 6, 2, 1).

In general, we say that a permutation has cycle-type 2r23r3 . . ., if it has exactly ri cyclesof length i, for i ≥ 2. So, for example, (1, 15)(2, 4, 6, 8, 7)(5, 9)(3, 11, 12, 13, 10)(14, 15, 16)has cycle-type 223152. By Proposition 5.7, conjugate permutations have the same cycle-type, and conversely, it is easy to see that if g and h have the same cycle-type, thenthere is an f ∈ Sym(X) with fgf−1 = h. For example, if X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},g = (1, 5, 9)(2, 4, 6, 8)(7, 10) and h = (1, 5)(2, 10, 9)(3, 6, 8, 7), then we can choose f to map1, 5, 9, 2, 4, 6, 8, 7, 10, 3 to 2, 10, 9, 3, 6, 8, 7, 1, 5, 4, respectively, so f = (1, 2, 3, 4, 6, 8, 7)(5, 10).(f is not unique; can you find some other possibilities?) Hence we have:

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Proposition 5.8 Two permutations of Sym(X) are conjugate in Sym(X) if and only if theyhave the same cycle-type.

For example, S3 has three conjugacy classes, corresponding to cycle-types 1, 21, 31, and S4

has five conjugacy classes, corresponding to cycle-types 1, 21, 22, 31, 41.

5.4 The Simplicity of A5

In Subsection 3.3, we defined a group G to be simple if its only normal subgroups are {1} andG, and we saw that the only abelian simple groups are the cyclic groups of prime order. Thereare also infinitely many finite non-abelian simple groups. These were eventually completelyclassified into a number of infinite families, together with 26 examples known as sporadicgroups, which do no belong to an infinite family. The work on this proof went on for decades,and was finally completed in about 1981.

One of the infinite families of finite non-abelian simple groups consists of the alternatinggroups An for n ≥ 5. The aim of this section will be to prove that A5 is simple. First westate an easy lemma.

Lemma 5.9 A subgroup H of a group G is normal in G if and only if H consists of a unionof conjugacy classes of G.

Proof: By Proposition 3.2, H �G if and only if ghg−1 ∈ H for all g ∈ G, h ∈ H. But thisis just saying that H �G if and only if h ∈ H =⇒ ClG(h) ⊂ H, and the result follows. 2

For example, consider the subgroup {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)} of S4. This is theunion of the conjugacy classes of S4 containing elements of cycle-types 1 and 22, and so it isa normal subgroup. Incidentally, since this subgroup lies in A4, it is also normal in A4, andso A4 is not simple.

We next need to find the conjugacy classes of A5. The classes of S5 correspond do cycle-types1, 21, 22, 31, 2131, 41, 51, and of these, the permutations of cycle-types 1, 22, 31 and 51 are evenpermutations and hence lie in A5.

There is 1 permutation of cycle-type 1, 15 of type 22, 20 of type 31, and 24 of type 51, making60 elements in total.

The problem is that these are classes in Sn, and two permutations could conceivably beconjugate in Sn but not in An, in which case the corresponding class would split up intomore than one conjugacy class in An.

In fact, the 15 permutations of cycle-type 22 forms a single class in An. To see this, letg = (x1, x2)(x3, x4)(x5) and h = (y1, y2)(y3, y4)(y5) be in this class. Then the permutationsf1 which maps xi to yi for 1 ≤ i ≤ 5 and f2 which maps x1 → y2, x2 → y1 and xi → yi for3 ≤ i ≤ 5 both satisfy fgf−1 = h, and since f2 = f1(x1, x2), one of f1 and f2 is an evenpermutation and lies in An. Hence g and h are conjugate in An.

Similarly, the 20 permutations of cycle-type 31 are all conjugate in An, because if g =(x1, x2, x3)(x4)(x5) and h = (y1, y2, y3)(y4)(y5), then f1 mapping xi → yi for 1 ≤ i ≤ 5and f2 mapping xi → yi for 1 ≤ i ≤ 3 and x4 → y5, x5 → y4 both satisfy fgf−1 = h.

However, for the cycle-type 51, if we take g = (1, 2, 3, 4, 5), h = (1, 2, 3, 5, 4), then the five per-mutations fi with figf

−1i = h are (4, 5)gk for 0 ≤ k ≤ 4, and these are all odd permutations,

so g and h are not conjugate in An. We can check that the 12 permutations (1, x2, x3, x4, x5)for which the map i→ xi (2 ≤ i ≤ 5) is an even permutation of {2, 3, 4, 5} are all conjugate tog in An, whereas the remaining 12, for which it is an odd permutation, are conjugate to h inAn. Hence this class in Sn splits into two conjugacy classes, each of size 12, in An. Summing

25

up, we have:

Lemma 5.10 A5 has 5 conjugacy classes, of sizes 1, 15, 20, 12, 12.

Theorem 5.11 A5 is a simple group.

Proof: By Lemma 5.9, a normal subgroup N of A5 would be a union of conjugacy classesof A5. But no combination of the numbers 1, 15, 20, 12, 12 that contains 1 adds up to adivisor of 60 other than 1 or 60, and so the result follows by Lagrange’s Theorem (2.10). 2

5.5 Sylow’s Theorem

By Lagrange’s Theorem (2.10), the order of a subgroup H of a finite group G always dividesthe order of G. An obvious converse question to ask is whether a group G has subgroups of allorders that divide |G|. This is true for some groups. For example, we saw in Subsection 2.1that it is true for finite cyclic groups. However, it is not true in general, and the smallestcounterexample is described in the following result:

Proposition 5.12 A4 has no subgroup of order 6.

Proof: |A4| = 24/2 = 12, so a subgroup H of order 6 would have index 2, and would benormal by Proposition 3.1, and hence by Lemma 5.9 a union of conjugacy classes of A4. ByProposition 3.5, a group of order 6 is cyclic or dihedral, and in either case it contains at leastone element of order 2. But the only elements of order two in A4 are (1, 2)(3, 4), (1, 3)(2, 4)and (1, 4)(2, 3), and these are all conjugate in A4 (check!), so they must all lie in H. But thenH has a subgroup of order 4, namely {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}, contradictingLagrange’s Theorem. 2

The Norwegian mathematician Sylow proved a number of theorems about subgroups of groupsof prime power order. We shall just prove the first one here, which asserts their existence.There are several proofs of this theorem. We have chosen one which demonstrates the use ofconjugacy classes, quotient groups and induction. We start with a lemma.

Lemma 5.13 Let G be an abelian group and let p be a prime dividing |G|. Then G has anelement of order p.

Proof: We use induction on |G|. The result is vacuously true when |G| = 1, so suppose that|G| > 1. Let g ∈ G with 1 6= g. By replacing g by a suitable power of itself if necessary, wemay assume that the order q of g is prime. If q = p then we are done. Otherwise, becauseG is abelian, the subgroup N of G generated by g is normal, and the quotient group G/Nhas order |G|/q, which is still divisible by p. Hence, by induction applied to G/N , G/N hasan element hN of order p. Then (hN)p = 1G/N = N , and so hp ∈ N . Either hp = 1, and|h| = p, or hp = gk for some k with 1 ≤ k < q, and then |h| = pq and |hq| = p. In either casewe have found an element of G of order p. 2

Theorem 5.14 Let G be a finite group and suppose that pn divides |G|, where p is primeand n > 0. Then G has a subgroup of order pn.

Proof: We use induction on |G|. The result is clear when |G| = 1, so suppose that |G| > 1.

Let the conjugacy classes of G be C1, C2, . . . , Cr, and let gi be an element in Ci. We assumethat C1 = {1} where g1 = 1. Then clearly

|C1|+ |C2|+ . . .+ |Cr| = |G| (†)

26

By Proposition 5.6, |Ci| = |G|/|CG(gi)|, and so |Ci| divides the order of |G|. Let us choosethe numbering such that |Ci| is not divisible by p for 1 ≤ i ≤ s, and |Ci| is divisible by p fors+ 1 ≤ i ≤ r. Since |G| is divisible by p, we must have s > 1, because if s = 1, then the leftand right hand sides of Equation (†) would be congruent to 1 and 0 mod p.

Suppose first that there is some Ci with 2 ≤ i ≤ s and |Ci| > 1. Then |CG(gi)| < |G| and soCG(gi) is a proper subgroup of G, and since p does not divide |Ci|, |CG(gi)| must be divisibleby pn. Then we can apply induction to CG(gi) and deduce that CG(gi) has a subgroup H oforder pn. But then H is also a subgroup of G, and we are done.

Hence we may assume that |Ci| = 1 for 1 ≤ i ≤ s. By Equation (†) again, we must have p|s.Now we saw above that |Ci| = 1 if and only if gi ∈ Z(G), so the centre Z(G) of G has order sand is equal to {g1, g2, . . . , gs}. Since p|s, it follows from the lemma above that Z(G) has anelement g of order p. Let N be the subgroup {1, g, g2, . . . , gp−1} of G generated by g. Sinceg ∈ Z(G), we have gi ∈ Z(G) for all i, and so hgih−1 = gi for all i and all h ∈ G, and N �G.

Now the quotient group G/N has order |G|/|N | = |G|/p, and by induction applied to G/Nwe know that G/N has a subgroup of order pn−1. By Proposition 4.6, such a subgroup hasthe form H/N where H is a subgroup of G of order |N ||H/N | = |N |pn−1 = pn, and we aredone. 2

Definition Suppose that n is the largest power of the prime p that divides |G|, so |G| = pnqwhere q is not divisible by p. A subgroup of G of order pn is called a Sylow p-subgroup of G.

We know from Theorem 5.14 that G has at least one Sylow p-subgroup. Here are the remain-ing theorems of Sylow, which we shall not prove here.

Theorem 5.15 Let G be a finite group, p a prime, and |G| = pnq, where p 6 | q. Then(i) The number r of Sylow p-subgroups of G satisfies r ≡ 1 (mod p);(ii) Any two Sylow p-subgroups of G are conjugate in G;(iii) Any subgroup of G of order pm for 1 ≤ m ≤ n is contained in a Sylow p-subgroup of G.

Appendix

Proof of Proposition 2.4: This proposition says that no permutation can be both evenand odd. We can assume that the set X being permuted is {1, 2, 3, . . . , n} for some n > 0.Let X{2} be the set of all unordered pairs {i, j} of elements of X. For an integer x, sgn(x) isdefined to be 1, 0, or −1, when x > 0, x = 0 and x < 0, respectively. For a permutation φ ofX, the sign sgn(φ) of φ is defined as:

sgn(φ) =∏

{i,j}∈X{2}

sgn(φ(j)− φ(i)

j − i

).

Let ψ = (k, l) be a transposition. We claim that sgn(φψ) = −sgn(φ). To see this, consider theeffect of replacing φ by φψ in the terms sgn

(φ(j)−φ(i)

j−i

)in the product above. If {k, l} ∩ {i, j}

is empty, then the term is unchanged. For those terms where {k, l} and {i, j} have one numberin common, say i = k, the product of the two terms sgn

(φ(j)−φ(k)

j−k

)and sgn

(φ(j)−φ(l)

j−l

)will be

unchanged. Finally the term sgn(

φ(k)−φ(l)k−l

)will change sign, so the whole product changes

sign, which establishes the claim.

Since the identity permutation has sign 1, it now follows that if φ is an even permutationthen sgn(φ) = 1, and if φ is an odd permutation then sgn(φ) = −1. The result follows. 2

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Part II

Ring Theory

6 Definition, Examples and Elementary Properties

6.1 Definitions

Definition A ring is a set R together with two binary operations+, · : R×R→ R that satisfy the following properties:(i) (Group under addition) (R,+) is an abelian group.(ii) (Associativity) For all a, b, c ∈ R, (ab)c = a(bc).(iii) (Distributivity) For all a, b, c ∈ R, (a+ b)c = ac+ bc and a(b+ c) = ab+ ac.(iv) (Identity) There exists an element 1 = 1R ∈ R such that, for all a ∈ R, 1a = a1 = a.

Note that we nearly always write ab in place of a · b. Some books omit Axiom (iv).

Lemma 6.1 Let R be a ring. Then R has a unique identity element.

Proof: Let 1 and 1′ be two identity elements of R. Then, 1 = 11′ = 1′. 2

As usual, the zero element of (R,+) will be denoted by 0R or usually just 0.

Lemma 6.2 In any ring R, 0a = a0 = 0 for all a ∈ R.

Proof: 0a = (0 + 0)a = 0a+ 0a, so 0a = 0 by cancellation in (R,+). Similarly a0 = 0. 2

What happens if 1 = 0? The proof of the following lemma is left as an exercise.

Lemma 6.3 Let R be a ring such that 0 = 1. Then R = {0}.

The ring {0} a ring is called the zero ring. All other rings will be called non-zero rings.

Definition A ring R is commutative if it satisfies(v) (Commutativity) For all a, b ∈ R, ab = ba.

Definition If a and b are non-zero elements of a ring R with ab = 0, then a and b are calledzero divisors. A non-zero commutative ring R is called an integral domain (or just domain)if it has no zero divisors; that is, if a, b ∈ R, ab = 0 implies a = 0 or b = 0.

Definition An element a of a ring R is called a unit if it has a two-sided inverse undermultiplication; that is, if there exists b ∈ R with ab = ba = 1.

Exercise Show that the units in R form a group under multiplication.

A non-zero ring R is called a division ring if R \ {0} is a group under multiplication; that is,if all of its non-zero elements are units. So a field is a commutative division ring.

Exercise Show that a field is a domain.

Proposition 6.4 (Cancellation laws in domains) If R is a domain, a, b, c ∈ R witha 6= 0, and ab = ac or ba = ca, then b = c.

Proof: ab = ac⇒ a(b− c) = 0 and since R is a domain with a 6= 0, this implies b− c = 0,so b = c. The proof for ba = ca is similar. 2

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Proposition 6.5 A finite integral domain is a field.

Proof: Let R = {r0, r1, . . . , rn} be a finite domain with 0 = r0. By Proposition 6.4, forfixed i > 0, the n products rirj (1 ≤ j ≤ n) are all distinct and non-zero. Since there aren possible values for these n products, they all occur exactly once. In particular, we haverirj = 1 for some j, so R is a field. 2

6.1.1 Examples – Numbers

Z, Q, R, and C are commutative rings under the usual addition and multiplication. Theyare all domains, and Q, R, and C are fields.

In Subsection 1.7 we saw that, for a fixed n > 0, Zn = {0, 1, . . . n − 1} forms a group underaddition modulo n. We can also define multiplication modulo n, and this makes Zn into acommutative ring. For instance, in Z6, 4 + 5 = 3 (residue of 9 modulo 6), 4 · 5 = 2 (residueof 20 modulo 6), 4 · 3 = 0 (residue of 12 modulo 6).

Proposition 6.6 Zn is a domain if and only if n is prime, in which case it is a field.

Proof: If n = 1 then Zn is the zero ring so is by definition not a domain. If n = ab with1 < a, b < n, then ab = 0 with a 6= 0 6= b in Zn, so Zn is not a domain. If n is prime and0 < a, b < n, then n does not divide ab (we shall prove this formally later), so ab 6= 0 in Zn,which is therefore a domain, and also a field by Proposition 6.5. 2

Exercise What are the units in Zn.

6.1.2 Examples – Matrices

If R is a ring then the set Mn(R) of n×n-matrices with coefficients in R is another ring. Themultiplication and addition is the same as you have learnt in linear algebra: (aij) + (bij) =(aij + bij) and (aij) · (bij) = (

∑k aikbkj).

For the zero ring R, the ring Mn(R) is also zero. For a non-zero ring Mn(R) is commutativeif and only if R is commutative and n = 1. The units in Mn(R) are of course the invertiblematrices.

6.1.3 Examples – Polynomials

Let R be a ring, and let x1, . . .xn be independent commuting variables. The polynomialsin xi-s with coefficients in R form the polynomial ring R[x1, . . . xn] under the usual additionand multiplication of polynomials.

A monomial is an expression xα = xα11 xα2

2 · · ·xαnn , where α1, . . . , αn are non-negative integers.

A polynomial is a linear combination of monomials with coefficients in R. For example,3x3

1x2/2 + 11x1 − 2x1x22/3 ∈ Q[x1, x2].

Observe that R[x1, . . . xn] is commutative if and only if R is commutative.

The most important example is the ring R[x] of polynomials in a single variable x. So eachf ∈ R[x] has the form f = anx

n + an−1xn−1 + · · · + a1x + a0, with ai ∈ R. If an 6= 0,

then we define n to be the degree deg(f) of f , and an is called the leading coefficient of f .A polynomial is called monic, if its leading coefficient is 1. Note that non-zero elements ofR have degree 0, and we leave the degree of the zero polynomial undefined, although someauthors define this to be −1 or −∞.

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6.2 Subrings

Definition A subset S of a ring R is called a subring of R if it forms a ring under the sameoperation as that of R with the same identity element.

Proposition 6.7 Let R be a ring and let S be a subgroup of (R,+). Then S is a subring ofR if and only if(i) a1, a2 ∈ S =⇒ a1a2 ∈ S; and(ii) 1R ∈ S.

Proof: As a subgroup of (R,+), (S,+) is an abelian group, closed under the multiplicationand containing the identity. Thus, S has two operations and identity for multiplication. Allring axioms of S easily follow from the corresponding axioms of R. 2

Lemma 6.8 The intersection of any set of subrings of R is itself a subring.

6.3 Isomorphisms and direct products

Definition An isomorphism φ : R → S between two rings R and S is a bijection from Rto S such that φ(r1r2) = φ(r1)φ(r2) and φ(r1 + r2) = φ(r1) + φ(r2) for all r1, r2 ∈ R. Tworings R and S are called isomorphic if there is an isomorphism between them. In this casewe write R ∼= S.

If φ : R→ S is an isomorphism, then we know from Lemma 4.1 for groups that φ(0R) = 0S .Since φ(1R) is an identity element of S, it follows from Lemma 6.1 that φ(1R) = 1S .

Definition Let R and S be two rings. We define the direct product R × S of R and S tobe the set { (r, s) | r ∈ R, s ∈ S } of ordered pairs of elements from R and S, with theobvious component-wise addition and multiplication (r1, s1) + (r2, s2) = (r1 + r2, s1 + s2),(r1, s1)(r2, s2) = (r1r2, s1s2) for r1, r2 ∈ R and s1, s2 ∈ S.

It is straightforward to check that R × S is a ring under these operations. Note that theidentity element is (1R, 1S).

But R and S are not isomorphic to subrings of R×S in general. Indeed, R can be thought ofas the elements of the form (r, 0S), and these elements define a ring isomorphic to R. But itsidentity element is (1R, 0S), which is not the same as the identity element (1R, 1S) of R× S.

Note also that the direct product of two non-zero rings is never a domain, since (1R, 0S)(0R, 1S) =(0R, 0S), the zero element of R× S.

Exercise (r, s) is a unit in R× S if and only if r is a unit in R and s is a unit in S.

In the proof of the next proposition, we shall use the following fact, which was proved lastyear in Foundations. We shall prove it again later on in the course. Integers m and n arecoprime (have no common prime divisors) if and only if there exist integers a and b such thatam+ bn = 1.

Proposition 6.9 (The Chinese Remainder Theorem) The rings Zm×Zn and Zmn areisomorphic if and only if m and n are coprime.

Proof: The “only if” bit follows from Proposition 1.10. If m and n are not relatively primethen their least common multiple l = lcm(m,n) < mn. The abelian additive groups are notisomorphic because the order of 1 is mn in (Zmn,+) but for any element (a, b) ∈ Zm × Zn,we have l(a, b) = (la, lb) = 0, so no element of Zm × Zn has order mn.

30

If m and n are coprime then there exist integers a, b such that am+ bn = 1. Let (x)n denotethe residue of x mod n. We define φ : Zmn → Zm × Zn by φ(x) = ((x)m, (x)n). It is clearthat φ(x+ y) = φ(x) + φ(y) and φ(xy) = φ(x)φ(y). It remains to show that φ is a bijection.We do it by writing the inverse map explicitly: φ−1(y, z) = (amz + bny)mn. Since both setshave the same cardinality mn, it suffices to observe that

φ(φ−1(y, z)) = φ((amz + bny)mn) = (((amz + bny)mn)m, ((amz + bny)mn)n) =

((amz + bny)m, (amz + bny)n) = ((bny)m, (amz)n) = (y, z)

with the last equality ensured by (bn)m = 1 = (am)n since am+ bn = 1. 2

Using induction on k, one derives the following corollary.

Corollary 6.10 If n = pa11 . . . pak

k is a decomposition of n into a product of distinct primesthen Zn

∼= Zpa11× · · · × Zp

akk

as rings.

7 Homomorphisms, Ideals, Quotient Rings and the

Isomorphism TheoremsThe theory of ring homomorphisms is very similar to that of group homomorphisms.

7.1 Homomorphisms

Definition Let R and S be rings. A ring homomorphism φ from R to S is a functionφ : R→ S such that φ(1R) = 1S , and for all r1, r2 ∈ R we have and φ(r1 +r2) = φ(r1)+φ(r2)φ(r1r2) = φ(r1)φ(r2).

From Lemma 4.1 for groups we have φ(0R) = 0S . Unlike in the case of a ring isomorphism,we cannot deduce that φ(1R) = 1S from the other conditions, so we choose to make it partof the definition (although in some books it is not!).

As for group homomorphisms, we define a ring monomorphism to be an injective homomor-phism and a ring epimorphism to be a surjective homomorphism. So it is an isomorphism ifand only if it is both a monomorphism and an epimorphism.

The image im(φ) of a ring homomorphism is just its image as a function, and it is straight-forward to check using Proposition 6.7 that it is a subring of S.

The kernel ker(φ) of a ring homomorphism is defined to be its kernel as a homomorphism ofadditive groups. That is,

ker(φ) = { r ∈ R | φ(r) = 0S }.

Examples

1. For a fixed integer n > 0, there is a ring homomorphism φ : Z → Zn in which, for k ∈ Z,φ(k) = k (mod n). That is, φ(k) is the residue of k modulo n. Then ker(φ) = nZ.

2. If φ : R→ S is a ring homomorphism, then there is an induced homomorphism ψ : R[x] →S[x], defined by

ψ(anxn + · · ·+ a1x+ a0) = φ(an)xn + · · ·+ φ(a1)x+ φ(a0).

Then ker(ψ) consists of those polynomial in which all coefficients lie in ker(φ).

Similalry, there is an induced homomorphism of matrix rings ψ : Mn(R) →Mn(S).

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3. Here is another example related to polynomials. Let R be a subring of S and let α bea fixed element of S. Then the map φ : R[x] → S defined by φ(f) = f(α), which evaluatesthe polynomial f at the point α, is a ring homomorphism in which ker(φ) consists of thosepolynomials with f(α) = 0.

4. Let R be a commutative ring of characteristic p, that is, px = 0 for any x ∈ R, where pis a prime number. The ring R admits a homomorphism, φ : R → R defined by φ(x) = xp.Clearly φ(xy) = φ(x)φ(y). The identity (x+y)p = xp +yp is sometimes called the Freshman’sdream binomial formula. It holds because the commutativity of x and y implies that

(x+ y)p = xp + yp +p−1∑k=1

p!k!(p− k)!

xkyp−k

and all binomial coefficients in the sum are divisible by p.

7.2 Ideals

Definition A subset I of a ring R is called an ideal in R if(i) I is a subgroup of (R,+);(ii) For all x ∈ R and y ∈ I, we have xy ∈ I and yx ∈ I.

Notice that ideals are different from subrings. They are both subgroups of (R,+), butsubrings satisfy Condition (ii) for ideals only when x and y are both in I. On the other hand,subrings must by definition contain 1R, whereas it is an easy exercise to show that an idealcontains 1R only when I = R.

When R is a commutative ring, it is routine to check that the subset (a) = { ra | r ∈ R },consisting of of all multiples of a in R, is an ideal of R, and it is known as the principal idealgenerated by a. This is also often written as aR or Ra.

For an arbitrary ring, the principal ideal (a) is equal to the set of finite sums {∑k

i=1 riasi |ri, si ∈ R }, but this will not be needed for this course.

Exercise If R is commutative, then (a) = R if and only if a is a unit of R.

In Z, the principal ideals are the sets (n) = nZ of multiples of n, for some fixed n ≥ 0. Weshall see later that these are the only ideals of Z.

In the theory of homomorphisms, ideals of rings correspond to normal subgroups of groups.

Proposition 7.1 Let φ : R→ S be a ring homomorphism. Then ker(φ) is an ideal in R.

Proof: By Theorem 4.4, K = ker(φ) is an additive subgroup of R. If r ∈ K, x ∈ R thenφ(xr) = φ(x)φ(r) = 0Sφ(r) = 0S . Hence xr ∈ K. Similarly, rx ∈ K and K is an ideal. 2

Exercise. Let φ : R → S be a ring homomorphism. Show that, if J is an ideal of S, thenφ−1(J) is an ideal of R. Give an example where I is an ideal of R but φ(I) is not an ideal ofS.

7.3 Quotient Rings

Since an ideal I of a ring R is a subgroup of (R,+), we can consider its cosets I+a for a ∈ R.We know from Theorem 3.9 that they form a group under addition, with the operation(I + a1) + (I + a2) = I + (a1 + a2). The following proposition defines the quotient ring R/I.

Proposition 7.2 The cosets of an ideal form a ring under addition in the quotient groupand the multiplication (I + a)(I + b) = I + ab.

32

Proof: We know already from group theory that R/I is a group under addition, so Axiom(i) of the definition of a ring is satisfied. We have to verify that the multiplication in R/I iswell-defined. To do this, suppose that I + a = I + x and I + b = I + y. So a− x, b− y ∈ I,and ab = ab− ay+ ay−xy+xy = a(b− y)+ (a−x)y+xy. But by the definition of an ideal,we have a(b− y) ∈ I and (a−x)y ∈ I, so ab−xy ∈ I and I+ab = I+xy, and multiplicationis well-defined. Axioms (ii), (iii) and (iv) follow immediately from the same properties in R,noting that 1R/I = I + 1R. 2

Examples 1. The quotient ring Z/(n) is isomorpic to the ring Zn of the residues modulon. The isomorphism Zn → Z/(n) is just m 7→ m + (n). This is an instance of the firstisomorphism for rings, which we shall prove shortly.

2. Let F be a field and f ∈ F [x] with deg(f) > 0. Let I = (f), the principal ideal of F [x]generated by f . Since (f) is not changed by multiplying f by a constant, we may assumethat f = xn + an−1x

n−1 + · · ·+ a1x+ a0 has leading coefficient 1.

Let us consider the quotient ring Q = F [x]/(f). The elements of Q are the cosets (f) + gwith g ∈ F [x]. Since f = xn + an−1x

n−1 + · · · + a1x + a0 ∈ (f), we can replace xn by−(an−1x

n−1+· · ·+a1x+a0) in g without changing the coset (f)+g. By doing this repeatedly,we can assume that deg(g) < n. For example, if f = x2 + 2x− 3, then

(f) + x3 = (f)− x(2x− 3) = (f) + 2(2x− 3) + 3x = (f) + 7x− 6.

On the other hand, if g, g′ ∈ F [x] with g 6= g′ and deg(g),deg(g′) < deg(f), then g−g′ 6∈ (f),so (f)+g and (f)+g′ are in distinct cosets of (f). Hence there is bijection between Q and thepolynomials of degree less than deg(f), where multiplication is done modulo f . In particular,if deg(f) = 1 then Q ∼= F .

Proposition 7.3 Let I be an ideal of a ring R. Then the map φ : R → R/I defined byφ(a) = I + a is a surjective ring homomorphism with kernel I.

Proof: It is straightforward to check that φ is a surjective ring homomorphism. Since0R/I = I, we have φ(a) = 0R/I ⇐⇒ I + a = I ⇐⇒ a ∈ I, so ker(φ) = I. 2

Theorem 7.4 (First Isomorphism Theorem for Rings) Let φ : R → S be a ring homo-morphism with kernel I. Then R/I ∼= im(φ). More precisely, there is an isomorphismφ : R/I → im(φ) defined by φ(I + a) = φ(a) for all a ∈ R.

Proof: By Theorem 4.7, φ is a well-defined isomorphism of abelian groups under addition.It is straightforward to check that it is a ring homomorphism. 2

7.4 Maximal ideals

Definition An ideal I of a ring R is called maximal, if I 6= R, but if J is any ideal of R withI ⊆ J ⊆ R, then I = J or J = R.

Theorem 7.5 An ideal I in a commutative ring R is maximal if and only if R/I is a field.

Proof: First suppose that I is maximal. Since I 6= R, 1 6∈ I, so I + 0 6= I + 1. We have tocheck that, for each x ∈ R \ I, (I+x) has a multiplicative inverse in R/I. Since I is maximaland x 6∈ I, the ideal I + (x) is equal to R, so 1 ∈ I + (x), and hence there exists y ∈ R with1 ∈ I + xy. Then I + 1 = I + xy = (I + x)(I + y).

Conversely, suppose that R/I is a field. Then I + 1 6= I + 0, so I 6= R. Let J be an ideal ofR with I ⊆ J ⊆ R. If I 6= J and x ∈ J \ I, then I + x 6= I + 0, so I + x has a multiplicativeinverse in R/I, and there exists y ∈ R with (I + x)(I + y) = I + 1. Hence I + 1 = I + xy, so1 ∈ I + (x) ⊆ J and thus J = R. 2

33

8 Euclidean Domains, Principal Ideal Domains, and

Unique Factorisation DomainsThe ring R will be an integral domain (and hence commutative) throughout this section.

8.1 Euclidean and Principal Ideal Domains

As you saw in Foundations, in the ring Z of integers, we have a Euclidean/division algorithm:

Proposition 8.1 For any a, b ∈ Z with 0 6= b, there exist q, r ∈ Z with a = qb + r, where0 ≤ r < |b|.

This can be proved by considering the least element r of the set {a− qb | q ∈ Z, a− qb ≥ 0}and showing that 0 ≤ r < |b|.There is a similar result for polynomial rings F [x] in one variable when F is a field.

Proposition 8.2 For any f, g ∈ F [x] with 0 6= g, there exist q, r ∈ F [x] with f = qg + r,where either r = 0 or deg(r) < deg(g).

Proof: If f = 0 we take q = r = 0, and otherwise we use induction on d = deg(f). If d = 0,then f ∈ F and then take q = 0, r = f if deg(g) > 0, and q = fg−1, r = 0 if deg(g) = 0.

So suppose that d > 0 and f = adxd+ad−1x

d−1+· · ·+a0. If deg(g) > d, then take q = 0, r = f .Otherwise g = bex

e + · · · + b0 with e = deg(g) ≤ d. Now deg(f − adb−1e xd−eg) < d and the

result follows by applying induction to f − adb−1e xd−eg. 2

These two examples motivate the following definition.

Definition A Euclidean domain (abbreviated ED) is a domain R that admits a norm functionν : R \ {0} → N ∪ {0} such that(i) ν(ab) ≥ ν(b) for all a, b ∈ R \ {0};(ii) ∀a, b ∈ R with b 6= 0, ∃ q, r ∈ R such that a = qb+ r and either r = 0 or ν(r) < ν(b).

Examples 1. Z and F [x] are Euclidean domains with ν(a) = |a| and ν(a) = deg(a),respectively.

2. Here is another example. Let i =√−1 and let Z[i] = {x + iy | x, y ∈ Z } be the ring of

Gaussian integers. It is straightforward to check using Proposition 6.7 that Z[i] is a subringof C, so it is certainly a domain. For z ∈ Z[i] define ν(z) = |z|2; that is, ν(x+ iy) = x2 + y2.Since ν(z1z2) = ν(z1)ν(z2), Condition (i) holds. Of course, we can define ν(z) in the sameway for all z ∈ C, and we still have ν(z1z2) = ν(z1)ν(z2).

Let a, b ∈ Z[i] with b 6= 0. Then we have a/b = x+ iy with x, y ∈ Q. Choose x0, y0 ∈ Z with|x− x0| ≤ 1/2 and |y − y0| ≤ 1/2. Then

a = b(x+ iy) = b(x0 + iy0) + b((x− x0) + i(y − y0)) = qb+ r,

where q = (x0 + iy0) ∈ Z[i] and r = b((x−x0)+ i(y−y0)). Since r = a− qb, we have r ∈ Z[i],and ν(r) = ν(b)ν((x− x0) + i(y− y0)) ≤ ν(b)(1/4 + 1/4) < ν(b), so Condition (ii) holds, andZ[i] is an ED.

We can prove various nice properties of EDs, but most of these properties hold in a moregeneral class of rings, which we shall now introduce. Recall that, in a commutative ring R,the principal ideals are those of the form (a) = aR for some fixed a ∈ R.

Definition A domain R is called a a principal ideal domain (abbreviated PID) if every idealof R is principal.

34

Theorem 8.3 Any Euclidean domain is a principal ideal domain. (ED =⇒ PID)

Proof: Let I be an ideal in a Euclidean domain R. If I = {0} then I = (0) is principal.Otherwise, choose b ∈ I \ {0} such that ν(b) is as small as possible. By definition of ideal,(b) ⊆ I. Let us now prove the opposite inclusion. For an arbitrary a ∈ I we can writea = bq + r with either r = 0 or ν(r) < ν(b). If r 6= 0 then r = a − bq ∈ I with ν(r) < ν(b),contradicting the choice of b. So a = bq ∈ (b) and I = (b) is principal. 2

Examples It turns out that the domain Z[α] with α = (1 +√−19)/2 is a PID but not a

ED, but we shall not prove that here. We shall however prove later that Z[α] with α =√−5

is not a PID.

Various familiar properties of divisibility that hold in Z hold in the more general context ofPIDs.

Definition Let x, y ∈ R we say that x divides y and write x|y if y = xr for some r ∈ R.

The following lemma is straightforward.

Lemma 8.4 The following statements are equivalent for all x, y ∈ R.(i) x|y;(ii) y ∈ (x);(iii) (x) ⊇ (y).

Definition Let x, y ∈ R. We say that x and y are associate (and write x ∼ y) if both x|yand y|x.

Lemma 8.5 The following statements are equivalent in a domain R:(i) x ∼ y;(ii) (y) = (x);(iii) There exists a unit q ∈ R with x = qy.

Proof: (i) ⇐⇒ (ii) follows from the previous lemma.(i) =⇒ (iii): For x = 0, we have x ∼ y ⇔ y = 0, so assume that x 6= 0 6= y. There existq, r ∈ R such that x = qy and y = rx. Then x = qy = q(rx) and (1 − qr)x = 0. Because Ris a domain, 1− qr = 0 and so q is a unit. (iii) =⇒ (i) is easy. 2

Examples In Z, the only units are ±1, so x ∼ y ⇐⇒ |x| = |y|.In F [x], with F a field, the units are the non-zero constants, so x ∼ y ⇐⇒ x = ay for somea ∈ F \ {0}. Note that every polynomial is associate to a unique monic polynomial (leadingcoefficient 1).

In Z[i], the multiplicative inverse of a+ ib in C is (a− ib)/(a2 + b2), which lies in Z[i] if andonly if a2 + b2 = 1. So there are four units, ±1 and ±i, and x ∼ y ⇐⇒ x = ±y or x = ±iy.

Definition Let x, y ∈ R. A greatest common divisor gcd(x, y) (also called highest commonfactor) is an element d ∈ R such that(i) d|x and d|y;(ii) if z ∈ R with z|x and z|y, then z|d.

A least common multiple lcm(x, y) is an element l ∈ R such that(i) x|l and y|l;(ii) if z ∈ R with x|z and y|z, then l|z.

35

This definition can be generalised in the obvious way to the gcd or lcm of any set of elementsof R.

Any two greatest common divisors gcd(x, y) of x and y must divide each other, and so areassociates. Similarly for lcms. So we have uniqueness up to associate elements. The sameholds for the gcd or lcm of a set of elements of R. We often write things like gcd(4, 6) = 2 tomean that 2 is a gcd of 4 and 6. This is a mild abuse of notation, since it is also true thatgcd(4, 6) = −2, but of course 2 6= −2. Note that gcd(0, x) = x and lcm(0, x) = 0 for anyx ∈ R.

Existence of greatest common divisors is a bit trickier. They do not necessarily exist in anarbitrary domain, but they do in a PID.

Proposition 8.6 If R is PID then lcm(x, y) and gcd(x, y) exist for any pair of elementsx, y ∈ R. Furthermore, there exist r, s ∈ R with gcd(x, y) = rx+ sy.

(We have already used this last property in Z in the proof of Proposition 6.9.)Proof: Let I = { rx+ sy | r, s ∈ R }. It is routine to check that I is an ideal of R (it is thesum (x) + (y) of the ideals (x) and (y)), so it is principal, and hence I = (d) for some d ∈ R.Similarly, the intersection (x) ∩ (y) is an ideal, so is equal to (l) for some l ∈ R. We claimthat d is a greatest common divisor and l is a least common multiple of x and y. Indeed,(x) ⊆ (d) ⊇ (y) and whenever (x) ⊆ (z) ⊇ (y) it follows that (z) ⊇ (x)+ (y) = (d). Similarly,(x) ⊇ (l) ⊆ (y) and whenever (x) ⊇ (z) ⊆ (y) it follows that (z) ⊆ (x) ∩ (y) = (l). 2

In an ED, we can compute r, s ∈ R with gcd(x, y) = rx+ sy by repeated application of theEuclidean algorithm. If ν(x) > ν(y) then we divide y into x and get a remainder x1, thendivide x1 into x to get a remainder x2 and carry on until the remainder is 0. At this stage,the last xi found is equal to gcd(x, y) (it is left as an exercise to prove this), and r and s canbe computed by substituting back into the equations found.

For example, in Z let x = 26, y = 6. Then

26 = 2× 10 + 610 = 1× 6 + 46 = 1× 4 + 24 = 2× 2 + 0

so gcd(26, 6) = 2, and substituting for 4 and then 6 gives 2 = 6−4 = 6−(10−6) = 2×6−10 =2× (26− 2× 10)− 10 = 2× 26− 5× 10.

8.2 Prime and Irreducible Elements

There are two different ways to say what a prime number is. We are going to see that theselead to two different notions in an arbitrary domain.

Definition Let r ∈ R \ {0}. We say that r is irreducible if r is not a unit, but r = ab witha, b ∈ R implies that either a or b is a unit.

We say that r ∈ R \ {0} is prime if r is not a unit, and r|xy implies that r|x or r|y.

Proposition 8.7 If R is a domain, then any prime element of R is irreducible.

Proof: Let r be prime and r = ab. Then, since r|r = ab, we have r|a or r|b. Withoutloss of generality, r|a. Since a|r, we have r ∼ a, so r = aq with q a unit. Then q = b byProposition 6.4, so b is a unit, and r is irreducible. 2

Proposition 8.8 If R is a PID, then any irreducible element of R is prime.

36

Proof: This one is a little tricky! Let r be irreducible and suppose that r|ab with a, b ∈ R.An element c = gcd(r, a) exists by Proposition 8.6. Then r = ct for some t ∈ R. Since r isirreducible, either c or t is a unit. We consider both cases.

If t is a unit, then r ∼ c and c|a, so r|a.So suppose that c is a unit. By Proposition 8.6, we have c = xa+ yr for some x, y ∈ R, andmultiplying by b gives cb = xab+ yrb. Now we know that r|ab and clearly r|yrb, so r|cb; thatis, ru = cb for some u ∈ R. But c a unit means that we can multiply by c−1 and concludethat r|b. 2

Example Let R = Z[√−5] = { a+ b

√−5 | a, b ∈ Z }. Then

6 = 2 · 3 = (1 +√−5)(1−

√−5)

in R. We claim that 2 is irreducible but not prime.

2 does not divide 1 ±√−5 because 2x = 1 ±

√−5 implies that x = 1/2 ±

√−5/2, which is

not an element of R. Hence, 2 is not prime.

Let us show that 2 is irreducible. If 2 = ab with a = x + y√−5, b = s + t

√−5 ∈ R then

4 = |a|2|b|2 = (x2+5y2)(s2+5t2). Clearly, |a|2, |b|2 ∈ N. If |a|2 = 1 then a−1 = x−y√−5 ∈ R

and a is a unit in R. Similarly, if |a|2 = 4 then |b|2 = 1 and b is a unit in R. Finally, it isclear that there are no integers x, y with x2 + 5y2 = 2, so we cannot have |a|2 = 2. Hence2 is irreducible as claimed. It can shown similarly that 3, (1 +

√−5) and (1−

√−5) are all

irreducible in R.

So by Proposition 8.8, we have

Corollary 8.9 Z[√−5] is not a PID and hence also not a ED.

8.3 Number Fields

Another interesting and easy property of PIDs is the following.

Proposition 8.10 For a 6= 0, the ideal (a) in a PID R is maximal if and only if a isirreducible.

Proof: If (a) is maximal then (a) 6= R, so a cannot be a unit. If a = bc, then (a) ⊆ (b) ⊆ R,and so either (a) = (b), in which case c is a unit, or (b) = R, in which case b is a unit. So ais irreducible.

Conversely, if a is irreducible, then a is not a unit, so (a) 6= R. Suppose (a) ⊆ (b) ⊆ R. Thenb|a, so a = bc for some c ∈ R and then either c is a unit, in which case (a) = (b), or b is aunit, in which case (b) = R. 2

In Example 2 in Subsection 7.3, we saw that, if F is a field and f ∈ F [x] with deg(f) > 0,then the elements of the quotient ring F [x]/(f) correspond to polynomials in F [x] of degreeless than f , where multiplication is done modulo f ; that is, f becomes equal to 0 in thequotient ring.

When f is irreducible, Proposition 8.10 and Theorem 7.5 together tell us that F [x]/(f) isa field. The case F = Q is particularly important, and as we are about to see, Q[x]/(f) isisomorphic to a subfield of C in this case.

Definition An element α ∈ C is said to be algebraic (over Q) if it satisfies a polynomialequation f(α) = 0 for some f ∈ Q[x] with deg(f) > 0. Otherwise α is called transcendental.

So, for example,√

2 is algebraic, and it has been proved that the well-known constants π ande are transcendental.

37

We saw in Example 3 of Subsection 7.1 that, for any α ∈ C, f(x) 7→ f(α) defines a ringhomomorphism φ : Q[x] → C. The property of α being transcendental is equivalent toker(φ) = {0}, in which case the First Isomorphism Theorem for Rings (Theorem 7.4) tells usthat im(φ) ∼= Q[x].

When α is algebraic, there exist non-zero f ∈ ker(φ) and, since ker(φ) is an ideal of thePID F [x], we have ker(φ) = (f) for some f ∈ F [x], where in fact we see from the proof ofTheorem 8.3 that f is a polynomial of smallest possible non-zero degree in ker(φ).

Although f is not unique, any two possible f would divide each other and thus be associates.By multiplying by a constant, we can assume that f has leading coefficient equal to 1, andthen it is unique, and is known as the minimal polynomial of α (over Q).

If f = gh with deg(g),deg(h) < deg(f) then f(α) = 0 implies g(α) = 0 or h(α) = 0,contradicting the fact that f has smallest possible degree in ker(φ). So f is irreducible.Summing up, we have proved:

Proposition 8.11 If α is an algebraic element of C, then there is a unique non-zero poly-nomial f ∈ Q[x] with leading coefficient 1 such that f(α) = 0, and f is irreducible.

By the First Isomorphism Theorem for Rings, we have im(φ) ∼= Q[x]/(f) which, as we sawabove, is a field, so im(φ) is a subfield of C, denoted by Q[α]. Fields of this type are callednumber fields and are the object of study of Algebraic Number Theory.

As a simple example, consider Q[√

2] = { a + b√

2 | a, b ∈ Q }, corresponding to f = x2 − 2.It is not completely obvious that this is a field, but this becomes clear from the identity

a+ b√

2c+ d

√2

=(ac− 2bd) + (bc− ad)

√2

c2 − 2d2,

which is derived by multiplying the top and bottom of the left hand side by c− d√

2.

8.4 Unique Factorisation Domains

Definition A domain R is a FD (factorisation domain) if each non-unit x ∈ R \ {0} admitsa factorisation x = r1r2 · · · rn where ri are irreducible elements.

An FD R is a UFD (unique factorisation domain) if for any two factorisations of an elementx = r1r2 · · · rn = s1s2 · · · sm, where all ri and si are irreducible, we have m = n and thereexists σ ∈ Sn such that ri ∼ sσ(i) for all i.

Proposition 8.12 Let R be a FD. Then R is a UFD if and only if every irreducible elementis prime.

Proof: For the only if part we consider an irreducible element x such that x|ab. Factorisinga = r1 · · · rk and b = rk+1 · · · rn, we get a factorisation ab = r1 · · · rn. On the other hand,ab = xy for some y ∈ R. Factorising y = s1 · · · st, we get another factorisation ab = xs1 · · · st.By the UFD property, x is associate to ri for some i. If i ≤ k then x|a. If i > k then x|b.For the if part, suppose that every irreducible element is prime, and consider two factorisa-tions x = r1r2 · · · rn = s1s2 · · · sm of a non-unit element x into irreducibles. We use inductionon n + m. If n = m = 1, then x = r1 = s1 and the result is clear. Otherwise, assumingwithout loss that n > 1, since rn|x and rn is prime, we have rn|si for some i with 1 ≤ i ≤ m.Hence, rnq = si, and irreducibility of si implies that q is a unit. Now the result follows byapplying the inductive assumption on r1r2 · · · rn−1 = (qs1) · · · si−1si+1 · · · sm. 2

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Theorem 8.13 A PID is a UFD. (So we have ED =⇒ PID =⇒ UFD.)

Proof: Using Propositions 8.8 and 8.12, it suffices to show that R is a FD.

We have to factorise an arbitrary non-unit x ∈ R \ {0} into irreducibles. Suppose there issuch an x that cannot be so factorised. So x is not irreducible, and we can write x = x1,1x1,2

where x1,i are not units. At least one of x1,1 and x1,2 is not irreducible, so we can factoriseit further.

We are going to repeat these steps over and over again. Step n + 1 starts with x =xn,1xn,2 · · ·xn,k where none of xn,i are units. They cannot all be irreducible, or we wouldhave successfully factorised x, so pick all of xn,i which are not irreducible, and write themas a product of two non-units resulting in x = xn+1,1xn+1,2 · · ·xn+1,t. This process goes onforever.

It is now more or less clear that we can find an infinite sequence of elements x = x0, x1, x2, . . .,where each xn = xn,i for some i, xi+1|xi but xi 6 |xi+1.

For those who prefer a more formal argument, we define a binary tree. The root of the treeis the element x. The nodes at level n are elements xn,i for all i. If xn,i is irreducible, it doesnot have any upward edges. If xn,i = xn+1,jxn+1,j+1, it has two upward edges going to xn+1,j

and xn+1,j+1. Since the process has not terminated, the tree is infinite. This means there isan infinite path in this tree starting from the root and going upward. Let xn = xn,i be theelement of this infinite path at level n. In particular, x0 = x.

We have done all this hard work to obtain the ascending chain of ideals

· · · ⊃ (xn+1) ⊃ (xn) ⊃ · · · ⊃ (x1) ⊃ (x0)

with all of the inclusions proper. Let I = ∪∞n=1(xn). Then I is an ideal. To see this, letr, s ∈ I. Then, for some m,n ≥ 0, we have r ∈ (xn), r ∈ (xm) and assuming without lossthat m ≥ n, we have r, s ∈ (xm), so r+ s ∈ (xm) ∈ I. The other conditions for being an idealcan be similarly checked.

Since R is a PID, I = (d) for some d ∈ R. Then d ∈ (yn) for some n. This implies thatI = (d) ⊆ (yn). But this contradicts the fact that (yn) is properly contained in (yn+1) ⊂ I,and this contradiction completes the proof. 2

Examples 1. All of our EDs Z, Z[i], F [X] (F a field) are UFDs.

2. Z[√−5] is a FD but not a UFD. We have seen that it is not a UFD since

6 = 2 · 3 = (1 + i√

5)(1− i√

5)

are two distinct factorisations into irreducibles. To prove it is an FD, use induction on |a|. If|a| = 1, then a is a unit (in fact the only units are ±1). If 0 6= a = bc with b and c non-units,0 < |b| < |a| and 0 < |c| < |a|, so by inductive hypothesis b and c factorise into irreducibles,and hence so does a.

3. Z[x] is a UFD, which will be proved later, but not a PID. The ideal

(2) + (x) = { 2a+ xb | a, b ∈ Z[x] }

is not principal.

Exercises Prove the assertion in Example 3 that (2) + (x) is not principal. Prove also that,for a field F and n > 1, the polynomial ring F [x1, . . . xn] is not a PID.

We saw in Proposition 8.6 that any two elements in a PID have a gcd and an lcm. We nowshow that the same holds in a UFD.

39

Proposition 8.14 Any finite collection of elements in a UFD has a gcd and an lcm.

Proof: Let a1, a2, . . . , ak be elements in a UFD R. Then each ai is divisible by only finitelymany irreducible elements, where we regard associate elements as being the same. Letp1, p2, . . . , pn be a complete list of the irreducible elements (or, more precisely, represen-tatives of their equivalence classes under associativity) dividing any of the ai. Then we canwrite each ai in the form uip

ei11 pei2

2 · · · peinn , where each eij ≥ 0 and ui is a unit. It is now not

hard to see that

gcd(a1, a2, . . . , ak) = pm11 pm2

2 · · · pmnn , and

lcm(a1, a2, . . . , ak) = pM11 pM2

2 · · · pMnn ,

where, for each j with 1 ≤ j ≤ n, mj = min(e1j , e2j , . . . , ekj) and Mj = max(e1j , e2j , . . . , ekj).2

8.5 Primes in F [x]

Throughout this section F [x] will be the polynomial ring over a field F . Since this is an ED,primes and irreducibles are the same. We start with the following easy observation.

Proposition 8.15 (Remainder Theorem) Let f = f(x) ∈ F [x]. Then for a ∈ F , f(a) = 0if and only if x− a divides f .

Proof: Since F [x] is an ED, we can divide f(x) by x− a with a remainder:

f(x) = g(x)(x− a) + r.

Since deg(x − a) = 1, we have r = 0 or deg(r) < 1, so r ∈ F (a constant polynomial).Substituting x = a, we arrive at f(a) = r, and the result follows. 2

Corollary 8.16 If 0 6= f ∈ F [x], then f(a) = 0 for at most deg(f) distinct values of a ∈ F .(In other words, a polynomial of degree d has at most d roots.)

Proof: Induction on deg(f). If deg(f) = 0 then, since f 6= 0, it is a non-zero constant, sof(a) is never 0. If deg(f) > 0 and f(a) = 0, then by Proposition 8.15, f = (x − a)g withdeg(g) = deg(f) − 1. Now, if f(b) = 0, then either b = a or g(b) = 0, and by inductivehypothesis there are at most deg(f)− 1 possible b with f(b) = 0, so the result follows. 2

Theorem 8.17 Let F be a field. Then any finite subgroup of the multiplicative group F \{0}is a cyclic group.

Proof: Suppose G ≤ F \ {0} is not cyclic, and let |G| = N . By the classification of finiteabelian groups (from Algebra 1), G is isomorphic to Cn1 × . . . × Cnm , a direct product ofcyclic groups of orders n1|n2 . . . |nm and N = n1n2 · · ·nm. Since G ≤ F \ {0} is not cyclic,m > 1. Now, putting n = nm, we have (x1, . . . , xm)n = (xn

1 , . . . , xnm) = (1, . . . , 1) for all

xi ∈ Cni , and so gn = 1 for any g ∈ G. But then, for f(x) = xn−1, there are N > n elementsa ∈ F with f(a) = 0, contradicting Corollary 8.16. 2

Using Proposition 6.6, we have the following corollary:

Corollary 8.18 If p is prime, then the multiplicative group Zp\{0} is a cyclic group of orderp− 1.

Definition A field F is algebraically closed if for any f(x) ∈ F [x] of degree at least 1 thereexists a ∈ F such that f(a) = 0.

40

Example C is the best known example of an algebraically closed field.

Another example is the subfield A of C consisting of the algebraic numbers, which are thenumbers a ∈ C with f(a) = 0 for some f ∈ Q[x]. (It is not obvious that A is a field, but itis!)

Proposition 8.19 If F is an algebraically closed field then the irreducibles in F [x] are thepolynomials of degree 1. So each irreducible is an associate of x− a for a unique a ∈ F .

Proof: The element x− a is irreducible because any of its divisors must have degree 1 or 0.If it is 0, the divisor is a unit. If it is 1, the divisor is associate to x− a.

To show that x − a and x − b are not associate when a 6= b, notice that since the units inF [x] are the non-zero constant polynomials, x− a is associate only to cx− ca for all c ∈ F ∗.

Finally, if we have an irreducible f ∈ F [x], then f has degree at least 1. Since F is alge-braically closed, there is a ∈ F such that f(a) = 0. By Proposition 8.15, x − a divides f .Hence f is associate to x− a and deg(f) = 1. 2

You should prove the next proposition yourself. It is an excellent exercise but we won’t usein this course.

Proposition 8.20 The monic primes in R[x] are x − a and x2 + bx + c for all possiblea, b, c ∈ R with b2 − 4c < 0.

It is much more difficult to test whether a polynomial in Q[x] is prime, and there is no easilystated necessary and sufficient condition for this. We shall see later in Theorem 9.5 thatthis problem is essentially equivalent to testing polynomials in Z[x] for irreducibility in Z[x].Factorising polynomials in Z[x] into irreducibles is one of the major research problems incomputer algebra.

8.6 Gaussian primes

In this section, we investigate which elements of Z[i] are prime. We will call them Gaussianprimes. Again Z[i] is an ED, so primes and irreducibles are the same.

Let us recall that ν(x) = |x|2 = xx∗, where x∗ is the complex conjugate of x. It is useful toremember that if x|y in Z[i] then ν(x)|ν(y) in Z.

Proposition 8.21 If x ∈ Z[i] and ν(x) is prime in Z then x is Gaussian prime.

Proof: Suppose y|x. Hence ν(y)|ν(x) = p in Z, which forces ν(y) to be p or 1. If ν(y) = pthen y is associate to x. If ν(y) = 1 then y is a unit. So x is irreducible and hence prime. 2

Proposition 8.22 Let p ∈ Z be a prime in Z. Then either p is a Gaussian prime or p = xx∗

where x and x∗ are Gaussian primes.

Proof: If p is not a Gaussian prime, then there exists a Gaussian prime x such that p = xyand neither x nor y is a unit. Hence, ν(x)ν(y) = ν(p) = p2. This forces ν(x) = ν(y) = p,which makes x and y prime by Proposition 8.21. Finally, xx∗ = ν(x) = p = xy, so x∗ = y. 2

Proposition 8.23 Let q ∈ Z[i] be a Gaussian prime. Then either ν(q) is a prime or a squareof a prime.

Proof: Let n = ν(q) = qq∗. Take the decomposition of n into primes in Z, say n = p1 · · · pt.Then q|pj in Z[i] for some j and hence ν(q) divides ν(pj) = p2

j . 2

41

Theorem 8.24 The prime elements in Z[i] are obtained from the prime elements Z. Eachprime p ∈ Z congruent to 3 modulo 4 is a Gaussian prime. The prime p = 2 gives rise to aGaussian prime q such that 2 = qq∗ ∼ q2. Each prime p ∈ Z congruent 1 modulo 4 gives riseto two conjugate Gaussian primes q and q∗ such that p = qq∗.

Proof: Let q be a Gaussian prime. Then, by Proposition 8.23, ν(q) = p or p2 for someprime p ∈ Z. In the first case (ν(q) = p), Proposition 8.22 tells us that p gives rise to twoGaussian primes q, q∗ such that p = qq∗ = ν(q).

In the second case (ν(q) = p2), since q is prime in Z[i] and q|p2, we have q|p in Z[i], andhence p = qs for some s ∈ Z[i]. Then |s| = |p|/|q| = ν(p)/ν(q) = p2/p2 = 1, so s is a unit(s−1 = s∗/|s|2). Hence q is associate to p. We saw earlier that the units in Z[i] are ±1 and±i, so q = ±p or ±ip in this case.

It remains to see that everything is controlled by the value of p modulo 4. If p = qq∗ andq = x + yi then p = qq∗ = x2 + y2. Since 12 = 1, 22 = 0, 32 = 1 in Z4, we cannot havex2 + y2 = 3 in Z4, and so a prime in Z congruent to 3 modulo 4 cannot be represented asqq∗. Thus, it must be prime in Z[i] as well.

For p = 2, we have 2 = (1− i)(1 + i) where, by Proposition 8.21, 1± i are Gaussian primes.Since (1− i) = −i(1 + i) and −i is a unit, (1− i) and (1 + i) are associates in Z[i], so 2 ∼ q2

with q = 1 + i.

Finally, we must prove that a prime p congruent to 1 modulo 4 is not a Gaussian prime. Wesaw in Proposition 8.18 that the multiplicative group Zp \ {0} is cyclic of order p − 1, andsince 4|p − 1, this group has an element a of order 4. Now the polynomial x2 − 1 of degree2 has at most 2 roots in Zp, and since 1 and −1(= p − 1) are roots, −1 is the only elementof order 2 in Zp \ {0}. Since a2 has order 2, we have a2 = −1 in Zp. So in Z, we have thata2 + 1 ≡ 0 (mod p), and there exists b ∈ Z with a2 + 1 = kp, so (a+ i)(a− i) = kp in Z[i].

If p were a Gaussian prime, then we would have p|a + i or p|a − i in Z[i], which is clearlyfalse, since multiples of p in Z[i] have the form px+ ipy for x, y ∈ Z. So p is not a Gaussianprime when p is congruent to 1 mod 4, which completes the proof. 2

Corollary 8.25 (Fermat) The prime 2 and every prime congruent to 1 modulo 4 is a sumof positive integer squares in a unique way.

Proof: If p = 2 or p is congruent to 1 mod 4 then by Theorem 8.24 we have p = qq∗ = a2+b2,where q = a + ib is a Gaussian prime. If p = a2 + b2 = c2 + d2 with a, b, c, d > 0, thenp = (a + ib)(a − ib) = (c + id)(c − id) are two prime decompositions in Z[i]. Since Z[i] is aUFD and the only units in Z[i] are ±1 and ±i, we have a+ib = ±(c±id) or a+ib = ±i(c±id)and hence a = c, b = d or a = d, b = c. 2

As an application, we will now determine which integers can be expressed as the sum of twointeger squares. So let S = {a2 + b2 | a, b ∈ Z}. Clearly c ≥ 0 for all n ∈ S.

Theorem 8.26 We have S = T , where T = {n2p1p2 · · · pk}, where n, k ∈ Z, k ≥ 0, and thepi are distinct positive primes in Z, each equal to 2 or congruent to 1 modulo 4.

Proof: First note that, for a2 + b2 and c2 + d2 in S, we have

(a2 + b2)(c2 + d2) = ν(a+ ib)ν(c+ id) = ν((a+ ib)(c+ id)) =

ν((ac− bd) + i(ad+ bc)) = (ac− bd)2 + (ad+ bc)2,

so S is closed under multiplication.

42

Clearly n2 ∈ S for any n ∈ Z, and if p > 0 is prime with pi = 2 or pi congruent to 1 mod 4then by Theorem 8.24 we have p = qq∗ = a2 + b2, where q = a + ib is a Gaussian prime, sop ∈ S. So T ⊆ S.

Assume that S 6= T and let c = a2 + b2 be the smallest element of S \ T . We can writeany positive integer as n2p1p2 · · · pk for distinct primes pi, so c 6∈ T implies that some pi iscongruent to 3 mod 4. Put p = pi. Then, by Theorem 8.24, p is a Gaussian prime dividinga2 + b2 = (a + ib)(a − ib), so p divides a + ib or a − ib. In either case, this implies p|a andp|b, so in fact p2|c. Since the primes pi are distinct, this forces p|n and hence p|n2. But then(a/p)2 + (b/p)2 = c/p2 = (n/p)2p1p2 · · · pk is a smaller element of S \ T . This contradictionshows that S = T . 2

Remark. Not all positive integers (for example 7) are the sum of three integer squares, butthere is a famous theorem of Lagrange that says that all positive integers can be expressedas the sum of four integer squares.

You should be able to prove for yourself that there are infinitely many primes in N that arecongruent to 3 modulo 4. Just adapt the proof that there are infinitely many primes. It isharder to prove that there are infinitely many primes congruent to 1 modulo 4, but we cando that as another application.

(There is a much more advanced result in Analytic Number Theory, due to Dirichlet, whichsays that any arithmetic progression { a+nd | n ∈ N } in which a and d are coprime integerscontains infinitely many primes.)

Corollary 8.27 There are infinitely many primes in N that are congruent to 1 modulo 4.

Proof: Suppose there are only finitely many, say p1, . . . , pn. Let p0 = 2, q0 = 1 + i,pj = qjq

∗j be a prime decomposition of pj . Let us consider a prime decomposition of x =

2p1p2 · · · pn + i ∈ Z[i]. No prime p ∈ Z congruent to 3 modulo 4 divides x, because x/p has1/p as the coefficient at i, so it is not in Z[i]. Hence, one of the Gaussian primes qj (if it is q∗jswap the notation between qj and q∗j ) divides x. So pj = ν(qj)|ν(x) = 4p2

1p22 · · · p2

n + 1, whichis a contradiction since ν(x) is congruent to 1 modulo all pj . 2

9 Polynomial Rings over UFDs

9.1 Fields of fractions

In Foundations, the construction of the field Q of rationals from the integers Z was describedin outline. We can perform this construction with any integral in place of Z. The details arestraightforward but unfortunately a little tedious!

Let R be a domain. We consider the set W = R × (R \ {0}) = { (x, y) ∈ R × R | y 6= 0 }.We define an equivalence relation on W by (a, b) ∼ (c, d) whenever ad = bc. It it left as anexercise to show that this is, indeed, an equivalence relation. An equivalence class of (a, b) iscalled a fraction and denoted a/b. Let Q = Q(R) be the set of all of the equivalence classeson W .

Proposition 9.1 If R is a domain then Q(R) is a field under the operationsa

b+c

d=ad+ bc

bd,

a

b· cd

=ac

bd,

and π : R→ Q(R), with π(r) = r/1 is an injective ring homomorphism

Proof: We have to show that these operations are well-defined. Then we have to establishall the axioms of a field. Finally we have to show that π is an injective ring homomorphism.

43

To show that the operations are well defined, we need to observe first that the denominatorsof the results are non-zero because R is a domain. It remains to prove that the result isindependent of the representative of the equivalence class. Given a/b = x/y and c/d = u/w,we need to show that ac/bd = xu/yw and (ad + bc)/bd = (xw + yu)/yw. The first equalityrequires acyw = bdxu, which follows directly from ay = bx and cw = du. The second equalityrequires adyw + bcyw = bdxw + bdyu. Rewriting it, we get adyw − bdxw = bdyu − bcyw,which holds because

adyw − bdxw = dw(ay − bx) = 0 and bdyu− bcyw = by(du− cw) = 0.

The list of axioms of a field is long and we have to go and check them all. But we are in agood shape because we know that the operations are well-defined, so we can use our usualintuition about fractions. The associativity of addition is probably the hardest axiom tocheck

(a

b+c

d) +

e

f=ad+ bc

bd+e

f=adf + (bcf + bde)

bdf=a

b+cf + de

df=a

b+ (

c

d+e

f).

The commutativity of addition is easier:

a

b+c

d=ad+ bc

bd=c

d+a

b.

The zero and the additive inverse are usual: 0 = 0/1 and −(a/b) = (−a)/b with all the checksroutine. The associativity of multiplication is straightforward

(a

b· cd) · ef

=ac

bd· ef

=ace

bdf=a

b· cedf

=a

b· cedf

=a

b· ( cd· ef

)

as well as the commutativity:a

b· cd

=ac

bd=c

d· ab.

The unity and the multiplicative inverse are usual: 1 = 1/1 and (a/b)−1 = b/a with all thechecks routine. It is worth noticing though why a 6= 0. Indeed, a = 0 if and only if a ·1 = b ·0if and only a/b = 0/1 = 0. Finally, we have to check distributivity but it suffices to do it onone side only because the multiplication is commutative:

(a

b+c

d) · ef

=ad+ bc

bd· ef

=ade+ bce

bdf=ade

bdf+bce

bdf=a

b· ef

+c

d· ef.

The map π is a ring homomorphism because π(1R) = 1/1 = 1Q,

π(xy) =xy

1=x

1· y1

= π(x) · π(y), π(x+ y) =x+ y

1=x

1+y

1= π(x) + π(y).

Finally, x is in the kernel if and only if x/1 = 0/1 if and only if x · 1 = 0 · 1 if and only ifx = 0. 2

Definition Q = Q(R) is called the field of fractions of a domain R.

Examples 1. Q(Z) = Q.

2. Q(F [x]) is equal to the field of rational functions p/q (p, q ∈ F [x], q 6= 0) in one variablex. This is commonly denoted by F (x) (so F [x] and F (x) are not the same!).

3. Q(Z[i]) = Q[i].

44

9.2 Polynomial rings over UFDs are UFDs

Our goal in this subsection is to prove that if R is a UFD then so is R[x].

If R is an integral domain and f, g ∈ R[x], then deg(fg) = deg(f) + deg(g), and hence:

Proposition 9.2 If R is an integral domain then so is R[x].

Another easy result is:

Proposition 9.3 If R is an integral domain then an irreducible element of R remains irre-ducible in R[x], and the units in R and in R[x] are the same.

Note that some of these properties can fail and give surprising results when R is not a UFD.For example, if R = Z4 (the ring of integers module 4) and f = 2x+ 1, we get f2 = 1, so fis a unit in R[x] \R. But here we shall confine our attention to domains R, for which thingsbehave more predictably.

For the remainder of this section, we shall assume that R is a UFD. We saw in Proposition 8.14that any finite set of elements in R has a gcd.

Definition An element 0 6= f = a0 + a1x + · · · + anxn ∈ R[x] is called primitive if

gcd(a0, a1, . . . , an) = 1.

So any non-zero f ∈ R[x] can be written as af0, where a ∈ R if the gcd of the coefficients off , and f0 is primitive.

Proposition 9.4 The product of two primitive polynomials is primitive.

Proof: Let f = a0 +a1x+ · · ·+anxn and g = b0 +b1x+ · · ·+bnxn be primitive, and suppose

that some irreducible p divides ci for 0 ≤ i ≤ m+ n, where fg = c0 + c1x+ · · ·+ cm+nxm+n.

Since f is primitive, p cannot divide every ai, so suppose that p|ai for 0 ≤ i < k, but p 6 | ak,where k ≥ 0. Similarly, choose l such that p|bi for 0 ≤ i < l, but p 6 | al, where l ≥ 0. We haveck+l =

∑k+li=0 aibk+l−i, where we take any undefined coefficients to be 0. Since p is prime that

does not divide ak or bl, it does not divide akbl, but p does divide every other term in thissum. So p 6 | ck+l, which is a contradiction. Hence fg must be primitive. 2

Theorem 9.5 Let R be a UFD with field of fractions Q = Q(R). Then a primitive polyno-mial in R[x] is irreducible if and only if it is irreducible in Q[x].

Proof: Let f ∈ R[x] be primitive. If deg(f) = 0 then f is a unit in R and so is irreducible inneither R[x] nor Q[x]. So assume that deg(f) > 0. Suppose that f = gh with g, h ∈ R[x] andg, h non-units. By primitivity, we cannot have deg(g) = 0 or deg(h) = 0, so if f is reduciblein R[x] then it is also reducible in Q[x].

Conversely, assume that f is primitive and reducible in Q[x], so f = gh with g, h ∈ Q[x]and g, h non-units – so deg(g),deg(h) > 0. Let a1 be the least common multiple of all thedenominators of the coefficients of g(x), so a1g ∈ R[x]. Now let a2 be the greatest commondivisor of all the coefficients of a1g(x), Put a = a1/a2. Then a ∈ Q and ag ∈ R[x] withag primitive. Similarly, we define b = b1/b2 ∈ Q with bh ∈ R[x] and bh primitive. So,by Proposition 9.4, f ′ := abgh = abf is primitive, and hence a2b2f

′ = a1b1f with f andf ′ both primitive. So a1b1 and a2b2 are both equal to the gcd of the coefficients of a1b1fand, by uniqueness of the the gcd, a1b1 and a2b2 are associates in R. But this means thatab = (a1b1)/(a2b2) is a unit u ∈ R, and then f = (ag)(u−1bh) is a factorisation of f in R[x],so f is reducible in R[x]. 2

45

In particular, we have Gauss’ Lemma, which says that a primitive irreducible polynomial inZ[x] remains irreducible in Q[x].

Corollary 9.6 If R is a UFD then there are two kinds of irreducibles in R[X]: irreducibleelements in R and primitive elements in R[X] that are irreducible in Q[X].

Proof: This follows from Proposition 9.3, Theorem 9.5, and the fact that a polynomial ofnon-zero degree in R[x] that is not primitive is reducible. 2

Theorem 9.7 If R is a UFD then so is R[x].

Proof: It is clear that we can factorise any f ∈ R[x] into a product of irreducible elementsof R and irreducible primitive polynomials in R[x]. For uniqueness, suppose that

p1p2 · · · pkf1f2 · · · fn = q1q2 · · · qlg1g2 · · · gm

be two factorisations of the same polynomial in R[x], where the pi and qi are irreducibles inR, and the fi and gi are irreducible primitive polynomials in R[x].

By Theorem 9.5, the fi and gi are also irreducible elements of Q[x] (with Q the field offractions of R), whereas the pi and qi are units in Q[x]. We saw earlier that Q[x] is an ED,so it is a UFD. Hence n = m and, after permuting the gi if necessary, fi and gi are associatesin Q[x] for 1 ≤ i ≤ n. This means that, for each i, we have gi = aifi for some ai ∈ Q. Butthen ai = bi/ci with bi, ci ∈ R, and cifi = bigi. Since fi and gi are primitive, ci and bi areboth gcds of the coefficients of cifi, so bi and ci are associates in R, and hence the ai are allunits in R.

We can now cancel the fi and conclude that p1p2 · · · pk = q1q2 · · · qla, where a = a1a2 · · · an

is a unit in R. It now follows from the fact that R is a UFD that k = l and, after permutingthe qi if necessary, pi and qi are associates for all i. So we have uniqueness. 2

Examples 1. Z[x] is a UFD. We saw earlier that Z[x] is not a PID.

2. By repeated application of Theorem 9.7, we find that R[x1, x2, . . . , xn] is a UFD for anyn > 0 and any UFD R.

As we remarked earlier, it is not easy to check polynomials in Z[x] for irreducibility. Thereis however a sufficient (but not necessary) condition, known as Eisenstein’s criterion that isoften useful. We shall state it for Z[x], but the same proof words for R[x] for any UFD R.

Proposition 9.8 Let f = a0 +a1x+ · · · anxn be a primitive polynomial in Z[x], and suppose

there is prime p ∈ Z such that p 6 | an, p|ai for 0 ≤ i < n and p2 6 | a0. Then f is irreducible inZ[x] (hence also in Q[x]).

Proof: Since f is primitive, if it is reducible in Z[x] then f = (bmxm+· · ·+b0)(clxl+· · ·+c0)with each bi, ci ∈ Z and 0 < l,m < n. Then a0 = b0c0 and, since p|a0 but p2 6 | a0, pmust divideone, but not both, of b0, c0. Suppose without loss that p|b0, p 6 | c0. Now a1 = b0c1+b1c0 so p|a1

implies p|b1c0, and since p 6 | c0, we get p|b1. Similarly, if n ≥ 3, then p|a2 = b0c2 + b1c1 + b2c0implies p|b2 and, since m < n, we can prove by induction that p|bm. But then p|an = bmcl,contrary to assumption. This contradiction proves that f is irreducible. 2

Examples 1. 2 + 12x+ 10x2 + 3x3 is irreducible using the prime 2.

2. We could replace the condition “p 6 | an, p|ai for 0 ≤ i < n and p2 6 | a0” by “p 6 | a0, p|ai for0 < i ≤ n and p2 6 | an” in Proposition 9.8, and the result would still be true, using a similarproof. So, for example 4− 9x+ 36x2 − 3x3 is irreducible, using the prime 3.

46

9.3 The cyclotomic polynomials

For a positive integer n, let ζn = e2πi/n. Then ζn is a complex n-th root of unity. The npowers 1 = ζ0

n, ζ1n, . . . , ζ

n−1n form all of the n-th roots of unity. These are distinct roots of

the polynomial xn − 1 and so in C[x] we have the factorisation

xn − 1 =n−1∏i=0

(x− ζin) (9.9)

Note that the set of n-th roots of unity forms a cyclic group under multiplication withgenerator ζn. An n-th root of unity is called primitive if it is not an m-th root of 1 for anym ∈ Z with 1 ≤ m < n. So the primitive roots are those of order n in the group; that is, thegenerators of the group.

Recall that gcd(0, n) = n for any n > 0.

Lemma 9.10 Let n > 0, and for 0 ≤ i < n define di = gcd(n, i). Then ζin has order n/di,

and so is a primitive (n/di)-th root of unity. In particular, ζin is a primitive n-th root of 1 if

and only if di = 1.

Proof: Since di|i, (ζin)n/di = (ζn

n )i/di = 1, so the order of ζin divides n/di. On the other

hand, if (ζin)m = 1 for some m > 0, then n divides im, so n/di divides m(i/di). But by the

definition of gcd, n/di and i/di are coprime, so n/di divides m. So the order of ζin is exactly

n/di as claimed. 2

For each integer n > 0, we define the n-th cyclotomic polynomial to be

Φn(x) =∏

0≤i<n

gcd(i,n)=1

(x− ζin)

So deg(Φn(x)) = φ(n), where φ is the Euler phi-function, defined as

φ(n) = { i | 0 ≤ i < n, gcd(i, n) = 1 }

for integers n > 0.

From the lemma above, we see that each power ζn is a d-th root of unity for some divisor d ofn. Conversely, for any d|n, the powers ζi

n of ζn with 0 ≤ i < n for which (n/d)|i are preciselythe the d-th roots of unity, so the sequence { ζi

n | 0 ≤ i < n } contains each d-th root of unityand hence each primitive d-th root of unity exactly once. So, from Equation 9.9, we have:

xn − 1 =∏d|n

Φd(x) (9.11)

This equation enables us to compute Φd(x) recursively. Here are the first few:

n 1 2 3 4 5 6Φd(x) x− 1 x+ 1 x2 + x+ 1 x2 + 1 x4 + x3 + x2 + x+ 1 x2 − x+ 1

n 7 8 9 10Φd(x) x6 + x5 + x4 + x3 + x2 + x+ 1 x4 + 1 x6 + x3 + 1 x4 − x3 + x2 − x+ 1

Observe, in particular that, for p prime,

Φp(x) = (xp − 1)/(x− 1) = xp−1 + xp−2 + · · ·+ x+ 1.

47

Proposition 9.12 For each n ≥ 0, Φn(x) is monic and has coefficients in Z.

Proof: Induction on n. Since Φ1(x) = x− 1, the result is true for d = 1, so suppose n > 1.Then by Equation 9.11, we have xn− 1 = f(x)Φn(x), where f(x) is the product of the Φd(x)over all divisors of n other than n itself. By inductive hypothesis, f(x) is monic and hascoefficients in Z. So Φn(x) is the result of dividing xn − 1 by f(x). By considering the usualprocess of dividing one polynomial by another and the fact that f(x) is monic, we see thatΦn(x) must also be monic have integers as coefficients. 2

It was at one time conjectured that the only coefficients that can occur in Φn(x) are 0, 1,and −1, but this turned out to be false. For example, Φ105(x) is:

x48 + x47 + x46 − x43 − x42 − 2x41 − x40 − x39 + x36 + x35 + x34 + x33 + x32 + x31 − x28 − x26

−x24 − x22 − x20 + x17 + x16 + x15 + x14 + x13 + x12 − x9 − x8 − 2x7 − x6 − x5 + x2 + x+ 1

Our final objective is to prove:

Theorem 9.13 Φn is irreducible in Z[x] for all n > 0.

For n prime, although Eisenstein’s irreducibility criterion does not apply directly to Φn(x),it can be applied to show that Φn(x + 1) and hence also Φn(x) is irreducible - this will bean exercise on an assignment sheet. The proof for general n is more difficult and will involvesome new ideas.

Although derivatives are only normally defined for functions over R or C, for an arbitraryfield F and f = anx

n + · · ·+ a1x+ a0 ∈ F [x], we can define the formal derivative Df of f tobe

nanxn−1 + (n− 1)an−1x

n−2 + · · ·+ 2a2x+ a1.

It can be checked that formal derivatives have many of the familiar properties of derivatives,including:(i) D(f+g) = Df + Dg(ii) D(af) = aD(f)(iii) D(fg) = fDg + gDf

for all f, g ∈ F [x] and a ∈ F .

((i) and (ii) are clear from the definition. To prove (iii), do it first for f = xn, g = xm andthen use (i), (ii) and f(g + h) = fg + fh for f, g, h ∈ F [x].)

In particular, note that if f = e2g with e, f, g ∈ F [x], then Df = 2geDe + e2Dg is divisibleby e.

Let p be a prime not dividing n. For a ∈ Z, we denote its residue modulo p in Zp by a.Then the map Z[x] → Zp[x] that maps anx

n + · · · + a1x + a0 to anxn + · · · + a1x + a0 is a

ring homomorphism (Example 2 in Subsection 7.1), and we denote the image of a polynomialf ∈ Z[x] under this map by f .

Lemma 9.14 If p is a prime not dividing n, and is as just defined, thengcd(xn − 1, Dxn − 1) = 1.

Proof: We have Dxn − 1 = nxn−1 and so xDxn − 1− nxn − 1 = n. But p 6 |n means that nis a unit in Zp, so we can multiply by n−1, and the result follows from Proposition 8.6. 2

Corollary 9.15 We cannot have e2|xn − 1 with e ∈ Zp[x] and deg(e) > 0.

48

Proof: As we saw above, this would imply that e was a common factor of xn − 1 andDxn − 1. 2

Lemma 9.16 (i) ap = a for all a ∈ Zp; (ii) f(xp) = f(x)p for all f ∈ Zp[x].

Proof: (i) Since the multiplicative group Zp \ {1} has order p− 1, we have ap−1 = 1 for alla ∈ Zp \ {1}, and hence ap = a. But this is also true for a = 0, so it is true for all a ∈ Zp.

(ii) Let f = anxn + · · ·+ a1x+ a0 ∈ Zp[x]. Recall from Example 4 of Subsection 7.1 that, for

a ring of characteristic p (that is, a ring in which px = 0 for all x ∈ R), the map x 7→ xp is aring homorphism of R. This applies to Zp[x], so f(x)p = ap

nxnp + · · · + ap1x

p + ap0, which by

(i) is equal to f(xp). 2

We shall also need the following easy lemma.

Lemma 9.17 Let F and K be fields with F ⊂ K, and let p, q be coprime polynomials inF [x]. Then p and q are also coprime as elements of K[x].

Proof: This follows from Proposition 8.6, since p, q coprime in F [x] implies that there existr, s ∈ F [x] with pr + qs = 1. But then, since r, s ∈ K[x], p, q are coprime in K[x]. 2

We turn now to the proof of Theorem 9.13. Suppose for a contradiction that Φn(x) is reduciblefor some n > 0, so we can write Φn(x) = f(x)g(x) with f, g ∈ Z[x] and deg(f),deg(g) > 0,where we may assume that f is irreducible and f(ζn) = 0. Since Φn is monic, we may assumethat f, g are monic.

Proposition 9.18 Let p be a prime not dividing n. Then f(ζpn) = 0.

Proof: Suppose not. Since p does not divide n, ζpn is a primitive n-th root of 1 and hence

Φn(ζpn) = 0. So f(ζp

n) 6= 0 implies g(ζpn) = 0, and hence ζn is a root of the polynomial

h(x) := g(xp). Since f and h have a common root ζn, they cannot be coprime in C[x], andso, by the lemma above, they are not coprime in Q[x]. But since f is irreducible in Q[x], wemust have gcd(f, h) = f , and hence f |h. Also, since f, h are monic, we have f |h in Z[x].

So with defined as above, we have, in Zp[x], Φn = fg, and f |h, where h(x) = g(xp) = g(x)p

by Lemma 9.16. So we cannot have gcd(f, g) = 1, since this would imply that gcd(f, h) = 1,contradicting f |h. So Φn = fg is divisible by e2, where e = gcd(f, g) = 1, with deg(e) > 1.But then e2|xn − 1, contradicting Corollary 9.15. 2

Corollary 9.19 Let 0 ≤ i < m with gcd(i, n) = 1. Then f(ζin) = 0.

Proof: We can write i = p1p2 . . . pr, where the pj are not necessarily distinct primes notdividing n. By Proposition 9.18, f(ζp1

n ) = 0. But we can apply it again with ζp1n in place of

ζn and conclude that f(ζp1p2n ) = 0. So an inductive proof shows that f(ζi

n) = 0. 2

We can now immediately complete the proof of Theorem 9.13. By this last corollary, f hasat least φ(n) distinct roots in C[x], and so deg(f) ≥ φ(n) = deg(Φn), which contradicts ourassumption that Φn = gh with deg(g),deg(h) > 0.

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